LIBRARY 

OF  THK 

University  of  California. 

©IFT    OF 

.WL'VL.''^.^ 

Class 

V 

\--' 


CHAUVENET'S 


TREATISE   ON 


Elementary  Geometry 


REVISED    AND    ABRIDGED 


BY 

W.  E.  BYERLY 

PROFESSOR  OF  MATHEMATICS  IN   HABVABD   ONIVERSITY 


PHILADELPHIA 

J.  B.  LIPPII^COTT    COMPANY 

1892. 


'aift^-^ 


as 


> 


Copyright,  1887,  by  J.  B.  LippIncott  Compant. 


3TER£OTYPERSAN0PRI!NT^s]ill' 


PREFACE. 


In  preparing  this  edition  of  Chauvenet's  Geometry 
I  have  endeavored  to  compel  the  student  to  think  and 
to  reason  for  himself,  and  I  have  tried  to  emphasize  the 
fact  that  he  should  not  merely  learn  to  understand  and 
demonstrate  a  few  set  propositions,  hut  that  he  should 
acquire  the  power  of  grasping  and  proving  any  simple 
geometrical  truth  that  may  he  set  before  him ;  and  this 
power,  it  must  be  remembered,  can  never  be  gained  by 
memorizing  demonstrations.  Systematic  practice  in  de- 
vising proofs  of  new  propositions  is  indispensable. 

On  this  account  the  demonstrations  of  the  main  propo- 
sitions, which  at  first  are  full  and  complete,  are  gradually 
more  and  more  condensed,  until  at  last  they  are  some- 
times reduced  to  mere  hints,  by  the  aid  of  which  the 
full  proof  is  to  be  developed;  and  numerous  additional 
theorems  and  problems  are  constantly  given  as  exercises 
for  practice  in  original  work. 

A  syllabus,  containing  the  axioms,  the  postulates,  and 
the  captions  of  the  main  theorems  and  corollaries,  has 
been  added  to  aid  student  and  teacher  in  reviews  and 
examinations,  and  to  make  the  preparation  of  new  proofs 
more  easy  and  definite. 


6  ELEMENTS   OF   GEOMETRY. 

In  the  order  of  the  propositions  I  have  departed  con- 
siderably from  the  larger  Chauvenet's  Geometry,  with 
the  double  object  of  simplifying  the  demonstrations 
and  of  giving  the  student,  as  soon  as  possible,  the  few 
theorems  which  are  the  tools  with  which  he  must  most 
frequently  work  in  geometrical  investigation. 

Teachers  are  strongly  advised  to  require  as  full  and 
formal  proofs  of  the  corollaries  and  exercises  as  of  the 
main  propositions,  and  to  lay  much  stress  upon  written 
demonstrations,  which  should  be  arranged  as  in  the 
illustrations  given  at  the  end  of  Book  I. 

In  preparing  a  written  exercise,  or  in  passing  a  written 
examination,  the  student  should  have  the  syllabus  before 
him,  and  may  then  conveniently  refer  to  the  propositions 
by  number.  In  oral  recitation,  however,  he  should  quote 
the  full  captions  of  the  theorems  on  which  he  bases  his 
proof. 

W.  E.  BYEELY. 

Cambridge,  Mass.,  1887. 


CONTENTS. 


PA  OB 

INTKODUCTION 9 

PLANE    GEOMETEY. 

BOOK   I. 

Rectilinear  Figures 12 

Exercises  on  Book  1 49 

BOOK    II. 
The  Circle.   Ratio.    Incommensurables.   Doctrine  of  Limits. 

Measure  of  Angles 55 

Exercises  on  Book  II 94 

BOOK    III. 

Theory  of  Proportion.    Proportional  Lines.    Similar  Poly- 
gons     101 

Exercises  on  Book  III 125 

BOOK    IV. 

Comparison  and  Measurement  of  the  Surfaces  of  Recti- 
linear Figures 130 

Exercises  on  Book  IV 146 

BOOK    V. 

Regular  Polygons.     Measurement  of  the  Circle 149 

Exercises  on  Book  V 169 

Miscellaneous  Exercises  on  Plane  Geometry 173 

Syllabus  of  Plane   Geometry.     Postulates,   Axioms,   and 

Theorems 182 

7 


8  CONTENTS.       , 

GEOMETEY    O^   SPACE. 
BOOK   VI. 

PAGK 

The  Plane.    Diedral  Angles.    Polyedral  Angles    ....    195 
Exercises  on  Book  VI 220 

BOOK   VII. 

POLYEDRONS 224 

Exercises  on  Book  VII 254 

BOOK   VIII. 

The  Three  Bound  Bodies.  The  Cylinder.  The  Cone. 
The  Sphere.  Spherical  Triangles.  Spherical  Poly- 
gons     257 

Exercises  on  Book  VIII 287 

BOOK    IX. 

Measurement  of  the  Three  Bound  Bodies 290 

Exercises  on  Book  IX 308 

Miscellaneous  Exercises  on  Solid  Geometry 309 

Syllabus  of  Propositions  in  Solid  Geometry.     Theorems  .    311 


ELEMENTS  OF  GEOMETRY. 


IKTEODUOTIOK 

1.  Every  person  possesses  a  conception  of  space  indefi- 
nitely extended  in  all  directions.  Material  bodies  occupy 
finite,  or  limited,  portions  of  space.  The  portion  of  space 
which  a  body  occupies  can  be  conceived  as  abstracted  from 
the  matter  of  which  the  body  is  composed,  and  is  called  a 
geometrical  solid.  The  material  body  filling  the  space  is 
called  a  physical  solid.  A  geometrical  solid  is,  therefore, 
merely  the  fornij  or  figure,  of  a  physical  solid.  In  this  work, 
since  only  geometrical  solids  will  be  considered,  we  shall,  for 
brevity,  call  them  simply  solids,  and  we  shall  define  them 
formally,  as  follows : 

Definition.  A  solid  is  a  limited  or  bounded  portion  of  space, 
and  has  length,  breadth,  and  thickness. 

2,  The  boundaries  of  a  solid  are  surface^. 

Definition.  A  surface  is  that  which  hae  length  and  breadth, 
but  no  thickness.    1 

If  a  surface  is  bounded,  its  boundaries  are  lir^es. 

If  two  surfaces  intersect,  their  intersection  is  a  line. 

Definition.  A  line  is  that  which  has  length,  but  neither 
breadth  nor  thickness. 

9 


10  ELEMENTS   OF   GEOMETRY. 

If  a  line  is  terminated,  it  is  terminated  by  points. 
If  two  lines  intersect,  they  intersect  in  a  point. 
Definition.  A  point  has  position,  but  neither  length,  breadth, 
nor  thickness. 

3.  If  we  suppose  a  point  to  move  in  space,  its  path  will 
be  a  line,  and  it  is  often  convenient  to  regard  a  line  as  the 
path,  or  locus,  of  a  moving  point. 

If  a  point  starts  to  move  from  a  given  position,  it  must 
move  in  some  definite  direction;  if  it  continues  to  move  in 
the  same  direction,  its  path  is  a  straight  line.  If  the  direc- 
tion in  which  the  point  moves  is  continually  changing,  the 
path  is  a  curved  line. 

If  a  point  moves  along  a  line,  it  is  said  to  describe  the 
line. 

By  the  direction  of  a  line  at  any  point  we  mean  the  direc- 
tion in  which  a  point  describing  the  line  is  moving  when  it 
passes  through  the  point  in  question. 

Definitions.  A  straight  line  is  a  line  which    -f f 

has  everywhere  the  same  direction. 

A  curved  line  is  one  no  portion  of 
which,  however  short,  is  straight. 

A  broken  line  is  a  line  composed  of 
different  successive  straight  lines. 

4.  Definitions.   A  plane   surface,   or 
simply  a  plane,  is  a  surface  such  that,  if 
any  two   points   in  it  are  joined  by  a 
straight  line,  the  line  will  lie  wholly  in 
the  surface. 

A  curved  surface  is  a  surface  no  portion  of  which,  however 
small,  is  plane. 

5.  Definitions.  A  geometrical  figure  is  any  combination  of 
points,  lines,  surfaces,  or  solids,  formed  under  given  condi- 


\  INTRODUCTION.  11 

tions.  Figures  formed  by  points  and  lines  in  a  plane  are 
called  plane  figures.  Those  formed  by  straight  lines  alone 
are  called  rectilinear^  or  right-lined^  figures;  a  straight  line 
being  often  called  a  right  line. 

6.  Definitions.  Geometry  may  be  defined  as  the  science  of 
extension  and  position.  More  specifically,  it  is  the  science 
which  treats  of  the  construction  of  figures  under  given  con- 
ditions, of  their  measurement  and  of  their  properties. 

Plane  geometry  treats  of  plane  figures. 
The  consideration  of  all  other  figures  belongs  to  the  geom- 
etry of  space,  also  called  the  geometry  of  three  dimensions. 

7.  Some  terms  of  frequent  use  in  geometry  are  here  de- 
fined. 

A  theorem  is  a  truth  requiring  demonstration.  A  lemma 
is  an  auxiliary  theorem  employed  in  the  demonstration  of 
another  theorem.  A  problem  is  a  question  proposed  for  solu^ 
tion.  An  axiom  is  a  truth  assumed  as  self-evident.  A  postu- 
late (in  geometry)  assumes  the  possibility  of  the  solution  of 
some  problem. 

Theorems,  problems,  axioms,  and  postulates  are  all  called 
propositions. 

A  corollary  is  an  immediate  consequence  deduced  from  one 
or  more  propositions.  A  scholium  is  a  remark  upon  one  or 
more  propositions,  pointing  out  their  use,  their  connection, 
their  limitation,  or  their  extension.  An  hypothesis  is  a  sup- 
position, made  either  in  the  enunciation  of  a  proposition  or 
in  the  course  of  a  demonstration. 


PLANE    GEOMETRY. 


BOOK  I 

RECTILINEAR  FIGURES. 
ANGLES. 

1.  Definition.  A  jplane  angle^  or  simply  an  angle,  is  the 
amount  of  divergence  of  two  lines  which  meet  in  a  point  or 
which  would  meet  if  produced  (i.e.,  prolonged). 

The  point  is  called  the  vertex  of  the  angle, 
and  the  two  lines  the  sides  of  the  angle. 

From  the  definition  it  is  clear  that  the  mag- 
nitude of  an  angle  is  independent  of  the  length  of  its  sides. 

An  isolated  angle  may  be  designated  by  the  letter  at  its 
vertex,  as  "  the  angle  O;"  but  when  several  angles  are  formed 
at  the  same  point  by  different  lines,  as  OA,  OB, 
OC,  we  designate  the  angle  intended  by  three  5 

letters;   namely,  by  one  letter  on  each  of  its         yC^'^^ 

sides,   together  with    the    one    at   its  vertex,      q— -i 

which  must  be  written  between  the  other  two. 

Thus,  with  these  lines  there  are  formed  three  different  angles, 

which  are  distinguished  as  AOB,  BOC,  and  AOG. 

Two  angles,  such  as  AOB,  BOO,  which   have  the   same 
vertex  0  and  a  common  side  OB  between  them,  are  called 
adjacent. 
12 


BOOK   I.  13 

2.  Definitions.  Two  angles  are  equal  when  one  can  be 
superposed  upon  the  other,  so  that  the  vertices  shall  co- 
incide and  the  sides  of  the  first  shall  fall  along  the  sides  of 
the  second. 

Two  angles  are  added  by  placing  them  in  the  same  plane 
with  their  vertices  together  and  a  side  in  common,  care  being 
taken  that  neither  of  the  angles  is  superposed  upon  the 
other.  The  angle  formed  by  the  exterior  sides  of  the  two 
angles  is  their  sum. 

3.  A  clear  notion  of  the  magnitude  of  an  angle  will  be 
obtained  by  supposing  that  one  of  its  sides,  as  OjB,  was  at 
first  coincident  with   the   other  side  OA^  and 

that  it  has  revolved  about  the  point  0  (turning 
upon  0  as  the  leg  of  a  pair  of  dividers  turns 
upon  its  hinge)  until  it  has  arrived  at  the  posi- 
tion OB.  During  this  revolution  the  movable  side  makes 
with  the  fixed  side  a  varying  angle,  which  increases  by  in- 
sensible degrees,  that  is,  continuously ;  and  the  revolving  line 
is  said  to  describe^  or  to  generate^  the  angle  AOB.  By  con-« 
tinning  the  revolution,  an  angle  of  any  magnitude  may  be 
generated. 

4.  Definitions.  When  one  straight  line  meets  another,  so  as 
to  make  two  adjacent  angles  equal,  each  of  these  angles  is 
called  a  right  angle;  and  the  first  line  is 

said  to  be  perpendicular  to  the  second. 

Thus,  if  ^ 00  and  BOC  are  equal  angles, 

each  is  a  right  angle,  and  the  line  CO  is      

perpendicular  to  AB. 

Intersecting  lines  not  perpendicular  are  said  to  be  oblique 
to  each  other. 

An  acute  angle  is  less  than  a  right  angle. 

An  obtuse  angle  is  greater  than  a  right  angle. 

2 


14  ELEMENTS   OF   GEOMETRY. 

5.  Definition.  Two  straight  lines  lying  in  the  same  plane 
and  forming  no  angle  with  each  other — that  is,  two  straight 
lines  in  the  same  plane  which  will  not  meet,  however  far 
produced — are  parallel. 


TKIANGLES. 

6.  Definitions.  A  plane  triangle  is  a  portion  of  a  plane 
bounded  by  three  intersecting  straight  lines;  as  A 50.  The 
sides  of  the  triangle  are  the  portions  of  the 
bounding  lines  included  between  the  points  of 
intersection;  viz.,  AB,  BC,  CA.  The  angles 
of  the  triangle  are  the  angles  formed  by  the 
sides  with  each  other;  viz.,  CAB,  ABC,  BCA.  The  three 
angular  points.  A,  B,  C,  which  are  the  vertices  of  the  angles, 
are  also  called  the  vertices  of  the  triangle. 

If  a  side  of  a  triangle  is  produced,  the  angle 
which  the  prolongation  makes  with  the  adja- 
cent side  is  called  an  exterior  angle ;  as  A  CD. 

A  triangle  is  called  scalene  (ABC)  when  no  two  of  its  sides 
are  equal ;  isosceles  (DEF)  when  two  of  its  sides  are  equal ; 
equilateral  {GUI)  when  its  three  sides  are  equal. 


A  right  triangle  is  one  which  has  a  right  angle ;  as  MNP, 
which  is  right-angled  at  ]V.  The  side  MP,  opposite  to  the 
right  angle,  is  called  the  hypotenuse. 

The  base  of  a  triangle  is  the  side  upon  which  it  is  supposed 


BOOK   I.  15 

to  stand.  In  general,  any  side  may  be  assumed  as  the  base ; 
but  in  an  isosceles  triangle  BEF^  whose  sides  DE  and  DF 
are  equal,  the  third  side  EF  is  always  called  the  base. 

When  any  side  ^(7  of  a  triangle  has  been  adopted  as  the 
base,  the  angle  BA  G  opposite  to  it  is*  called 
the  vertical  angle,  and  its  angular  point  A 
the  vertex  of  the  triangle.  The  perpen- 
dicular AD  let  fall  from  the  vertex  upon 
the  base  is  then  called  the  altitude  of  the  triangle. 

7.  Definition.  Equal  figures  are  figures  which  can  be  made 
to  coincide  throughout  if  one  is  properly  superposed  upon 
the  other. 

Eoughly  speaking,  equal  figures  are  figures  of  the  same 
size  and  of  the  same  shape ;  equivalent  figures  are  of  the 
same  size  but  not  of  the  same  shape;  and  similar  figures 
are  of  the  same  shape  but  not  of  the  same  size. 

POSTULATES  AND  AXIOMS. 

8.  Postulate  I.  Through  any  two  given  points  one  straight 
line,  and  only  one,  can  be  drawn. 

Postulate  II.  Through  a  given  point  one  straight  line,  and 
only  one,  can  be  drawn  having  any  given  direction. 

9.  Axiom  I.  A  straight  line  is  the  shortest  line  that  can 
be  drawn  between  two  points. 

Axiom  II.  Parallel  lines  have  the  same  direction. 


16  ELEMENTS   OF   GEOMETEY. 

PROPOSITION  I.— THEOREM. 

10.  At  a  given  point  in  a  straight  line  one  perpendicular  to 
the  line  can  be  drawn,  and  but  one. 

Let  0  be  the  given  point  in  the  line  AB. 

Suppose  a  line  OD,  constantly  passing  through  0,  to 
revolve  about  0,  starting  from  the  position  OA  and  stopping 
at  the  position  OB. 

The  angle  which  01)  makes  with  OA  will  at  first  be  less 
than  the  angle  which  it  makes  with  OB,  and  will  eventually 
become  greater  than  the  angle  made  with 
OB.  ^        D 

Since  the  angle  DOA  increases  continu-                   j  / 
ously  (3),  the  line  OB  must  pass  through     ]/_ 

...  BOA 

one  position  m  which  the  angles  DOA  and 

DOB   are   equal.     Let   OC  he   this   position.     Then  0(7  is 

perpendicular  to  ^jB  by  (4).* 

There  can  be  no  other  perpendicular  to  AB  at  0,  for  if 
OD  is  revolved  from  the  position  0(7  by  the  slightest  amount 
in  either  direction,  one  of  the  adjacent  angles  will  be  in- 
creased at  the  expense  of  the  other,  and  they  will  cease  to 
be  equal. 

11.  Corollary.  Through  the  vertex  of  any  given  angle  one 
line  can  be  drawn  bisecting  the  angle,  and  but  one. 

Suggestion.  Suppose  a  line  OD  to  revolve 
about  O,  as  in  the  proof  just  given. 


*  An  Arabic  numeral  alone  refers  to  an  article  in  the  same  Book ;  but 
in  referring  to  articles  in  another  Book,  the  number  of  the  Book  is  also 
given. 


V, 


BOOK   I. 


17 


PROPOSITION  II.— THEOREM. 

12.  All  right  angles  are  equal. 

Let  AOG  and  AIQfC  be  any  two  right  angles. 

Superpose  A'CfC  upon  AOC^ 
placing  the  point  (J  upon  the 
point  0  and  making  the  line  C/A' 
fall  along  the  line  OA;  then  will 
OC  coincide  with  00,  for  other-  5 — 
wise  we  should  have  two  perpen- 
diculars to  the  line  AB  at  the  same  point  0,  which,  by  Propo- 
sition I.,  is  impossible. 

The  angles  A'C/C  and  AOC  are  then  equal  by  definition  (2). 


B' 


(y 


PROPOSITION  III.— THEOREM. 

13.  The  two  adjacent  angles  which  one  straight  line  makes 
with  another  are  together  equal  to  two  right  angles. 

If  the  two  angles  are  equal,  they  are  right  angles  by 
definition  (4),  and  no  proof  is  necessary. 

If  they  are  not  equal,  as  AOD  and  BOD,  still  the  sum  of 
AOD  and  BOB  is  equal  to  two  right  angles. 

Let  OC  be  drawn  at  0  perpendicular  to 
AB. 

The  angle  AOD  is  the  sum  of  the  two 
angles  AOC  and   COD  (2).    Adding  the 
angle  BOD,  the   sum  of  the  two   angles 
AOD  and  BOD  is  the  sum  of  the  three  angles  AOC,  COD, 
and  BOD. 

The  first  of  these  three  is  a  right  angle,  and  the  other  two 
are  together  equal  to  the  right  angle  BOC ;  hence  the  sum 
of  the  angles  AOD  and  BOD  is  equal  to  two  right  angles. 


18 


ELEMENTS   OF   GEOMETRY. 


14.  Corollary  I.    The  sum  of  all 

angles  having  a  common  vertex^  and 
formed  on  one  side  of  a  straight  line,     ^ 
is  two  right  angles. 


15.  Corollary  II.  The  sum  of  all 

the  angles  that  can  be  formed  about 

a  point    in    a  plane    is  four  right 
angles. 


EXERCISE. 

Theorem. — If  a  line  is  perpendicular  to  a  second  line,  then 
reciprocally  the  second  line  is  perpendicular  to  the  first. 

16.  Definition.  When  the  sum  of  two  angles  is  equal  to  a 
right  angle,  each  is  called  the  complement 
of  the  other.     Thus,  J)  0(7  is  the  comple- 
ment of  A  ODj  and  A  OD  is  the  complement 
of  DOC. 

When  the  sum  of  two  angles  is  equal  to     b 
two  right  angles,  each  is  called  the  supple- 
ment of  the  other.     Thus,  BOD  is  the  supplement  of  AOD, 
and  AOD  is  the  supplement  of  BOD. 

It  is  evident  that  the  complements  of  equal  angles  are 
equal  to  each  other ;  and  also  that  the  supplements  of  equal 
angles  are  equal  to  each  other. 


BOOK   I.  19 

PROPOSITION  IV.— THEOREM 

17.  If  the  sum  of  two  adjacent  angles  is  equal  to  two  right 
angles^  their  exterior  sides  are  in  the  same  straight  line. 

Let  the  sum  of  the  adjacent  angles  AOD^  BOD,  be  equal 
to  two  right  angles ;  then  OA  and  OB  are 
in  the  same  straight  line. 

For  BOD  is  the   supplement  of  A 02),     

and  is  therefore  identical  with  the  angle 

which  OD  makes  with  the  prolongation  of  ^  0  (Proposition 

TIL). 

Therefore  OB  and  the  prolongation  of  AO  are  the  same 
line. 

18.  Every  proposition  consists  of  an  hypothesis  and  a  con- 
clusion. The  converse  of  a  proposition  is  a  second  proposition 
of  which  the  hypothesis  and  conclusion  are  respectively  the 
conclusion  and  hypothesis  of  the  first.  For  example,  PropO' 
sition  III.  may  be  enunciated  thus : 

Hypothesis — if  two  adjacent  angles  have  their  exterior 
sides  in  the  same  straight  line,  then — Conclusion — the  sum 
of  these  adjacent  angles  is  equal  to  two  right  angles. 

And  Proposition  TV.  may  be  enunciated  thus : 

Hypothesis — if  the  sum  of  two  adjacent  angles  is  equal  to 
two  right  angles,  then — Conclusion — these  adjacent  angles 
have  their  exterior  sides  in  the  same  straight  line. 

Each  of  these  propositions  is,  therefore,  the  converse  of  the 
other. 

A  proposition  and  its  converse  are,  however,  not  always 
both  true. 


20  ELEMENTS   OF   GEOMETRY. 

PROPOSITION  v.— THEOREM. 

19.  If  two  straight  lines  intersect  each  other,  the  opposite  (or 
vertical)  angles  are  equal. 

B  0 

Let  AB  and  CD  intersect  in  0 ;  then  will 
the  opposite,  or  vertical,  angles  AOG  and 
BOD  be  equal. 

For  each  of  these  angles  is,  by  Proposi- 
tion III.,  the  supplement  of  the  same  angle  BOC,  and  hence, 
they  are  equal  (16). 

In  like  manner  it  can  be  proved  that  the  opposite  angles 
AOD  and  BOC  are  equal. 

EXERCISES. 

1.  Theorem. — The  line  which  bisects  one  of  two  vertical  angles 
bisects  the  other. 

2.  Theorem. — The  straight  lines  which 
bisect  a  pair  of  adjacent  angles  formed 
by  two  intersecting  straight  lines  are  per- 
pendicular to  each  other. 

Suggestion.  Prove  EOH  =  FOH. 

Q 

PROPOSITION  VI.— THEOREM. 

20.  Two  triangles  are  equal  when  two  sides  and  the  included 
angle  of  the  one  are  respectively  equal  to  two  sides  and  the 
included  angle  of  the  other. 

In  the  triangles  ABC,  DEF,  let  AB  be  equal  to  DE,  BC  to 
EF,  and  the  included 

'  AD  D 

angle  B  equal  to  the 
included  angle  E;  then 
the  triangles  are  equal. 
For,    superpose    the 


.X* 


B  C     E  F  F  E 

triangle  ABC  upon  the  triangle  DEF,  placing  the  point  B 


BOOK   I.  21 

upon  the  point  E,  and  making  the  side  BC  fall  along  the 
side  EF.  Then,  since  J5(7  is  equal  to  EF  by  hypothesis,  the 
point  G  will  fall  upon  the  point  F. 

Since  the  angle  B  is  equal  to  the  angle  E^  and  the  side 
BG  has  been  made  to  coincide  with  the  side  EF^  BA  must 
fall  along  ED^  by  definition  (2) ;  and,  as  BA  is  equal  to  ED^ 
the  point  A  will  fall  on  the  point  D. 

Since  the  point  G  has  been  proved  to  coincide  with  the 
point  F^  and  the  point  A  with  the  point  D,  the  side  GA  must 
coincide  with  the  side  FD^  by  Postulate  I.  (8). 

The  two  triangles  have  now  been  proved  to  coincide 
throughout,  and  are  equal,  by  definition  (7). 

21.  Scholium.  When  two  triangles  are  equal,  the  equal 
angles  are  opposite  to  the  equal  sides. 

PROPOSITION  VII.— THEOREM. 

22.  Two  triangles  are  equal  when  a  side  and  the  two  adjacent 
angles  of  the  one  are  respectively  equal  to  a  side  and  the  two 
adjacent  angles  of  the  other. 

In  the  triangles  ABG,  DEF,  let  BG  be  equal  to  EF,  and 
let  the  angles  B  and  G  adjacent  to  BGhe  respectively  equal 
to  the  angles  E  and  F 
adjacent  to  EF;  then 
the  triangles  are  equal. 

For,    superpose    the  \       \  \       \    ..--'''       / 

..1.^^  ,  B  C    E  F  F  E 

triangle  ABG  upon  the 

triangle  BEF,  placing  the  point  B  upon  the  point  E,  and 

making  the  side  BG  fall  along  the  side  EF. 

Since  J5(7  is  equal  to  EF,  the  point  G  will  fall  upon  the 
point  F. 

Since  the  angle  B  is  equal  to  the  angle  E,  by  hypothesis, 


22  ELEMENTS   OF   GEOMETRY. 

and  the  side  5(7  has  been  made  to  coincide  with  the  side  EF^ 
BA  must  fall  along  ED^  and  the  point  A  will  fall  somewhere 
on  the  side  ED,  or  on  that  side  extended. 

Since  the  angle  G  is  equal  to  the  angle  P,  by  hypothesis, 
and  BC  coincides  with  EF^  CA  must  fall  along  FD,  and  the 
point  A  will  fall  somewhere  on  the  line  FD,  or  on  that  line 
extended. 

Since  A  has  been  proved  to  lie  upon  ED,  and  also  upon 
FD,  it  must  coincide  with  the  only  point  they  have  in  com- 
mon, the  point  D. 

Hence  the  triangles  coincide  throughout,  and  are  equal. 

PEOPOSITION  VIII.— THEOREM. 

23.  In  an  isosceles  triangle  the  angles  opposite  the  equal  sides 
are  equal. 

Let  ^5  and  AC  he  the  equal  sides  of  the  isosceles  triangle 
ABC;  then  the  angles  B  and  C  are  equal. 

Through  the  vertex  A  draw  a  line  AD,  bisect- 
ing the  angle  BAC,  and  meeting  the  side  BC  t 
atD.                                                                                 / 

In  the  triangles  ABD  and  ACD  the  side  AB 
is  equal  to  the  side  AChy  hypothesis,  the  side 
AD  is  common,  and  the  included  angle  BAD  is  equal  to 
the  included  angle  CAD  by  construction.     The  triangles  are 
therefore  equal,  by  Proposition  YI.,  and  the  angle  C  of  the 
one  is  equal  to  the  angle  B  of  the  other,  by  (21). 

24.  Corollary.  The  straight  line  bisecting  the  vertical  angle 
of  an  isosceles  triangle  bisects  the  base,  and  is  perpendicular  to 
the  base. 

EXERCISE. 

Theorem. — An  equilateral  triangle  is  also  equiangular. 


BOOK  I.  23 

PROPOSITION  IX.— THEOREM. 

25.  Two  triangles  are  equal  when  the  three  sides  of  the  one 
are  respectively  equal  to  the  three  sides  of  the  other. 

In  the  triangles  ABC,  DEF,  let  AB  be  equal  to  BE,  AG 
to  DF,  and  BG  to  EF ;  then 
the  triangles  are  equal.  ^ 

For,    suppose    the     triangle      ^^     \  ^^ 

ABG  to  be  placed  so  that  its     ^  ^         \^^ 

base  ^(7  coincides  with  its  equal 
EF,  but  so  that  the  vertex  A 

falls  on  the  opposite  side  of  EF  from  X),  as  at  G,  and  join 
D  and  6^  by  a  straight  line. 

The  triangle  EBG  is  isosceles,  since  the  side  ED  is  equal 
to  the  side  EG  by  hypothesis;  therefore  the  angles  EBG 
and  EQD  are  equal,  by  Proposition  YIII. 

The  triangle  FBG  is  isosceles,  since  the  side  FI)  is  equal 
to  the  side  FG  by  hypothesis;  therefore  the  angles  FDG 
and  FGD  are  equal,  by  Proposition  YIII. 

If  to  the  equal  angles  EDG  and  EGB  we  add  the  equal 
angles  FBG  and  FGB,  the  sums  will  be  equal,  and  we  have 
the  whole  angle  EBF  equal  to  the  whole  angle  EGF. 

The  two  triangles  EBF  and  EGF  have  now  the  side  EB 
equal  to  the  side  EG  by  hypothesis,  the  side  BF  equal  to 
the  side  FG  by  hypothesis,  and  the  included  angle  EBF 
proved  equal  to  the  included  angle  EGF.  Hence  the  tri- 
angles are  equal,  by  Proposition  YI. 

EXERCISE. 

Theorem. — A  line  drawn  from  the  vertex  of  an  isosceles  tri- 
angle to  the  middle  point  of  the  base  is  perpendicular  to  the  base, 
and  bisects  the  vertical  angle. 


24  ELEMENTS   OF   GEOMETRY. 

PROPOSITION  X.— THEOREM. 

26.  Two  right  triangles  are  equal  when  they  have  the  hypote- 
nuse and  a  side  of  the  one  respectively  equal  to  the  hypotenuse 
and  a  side  of  the  other. 

In  the  right  triangles  ABC,  A'B'G\  let  the  hypotenuse  AB 
be  equal  to  the  hypotenuse 
A'B',  and  the  side   ^C  be 
equal  to  the  side  BC^ ;  then 
the  triangles  are  equal. 

Extend  the  side  BC  to  D, 
making  CD  equal  to  BC,  and 

join  A  and  D ;  and  extend  B'C^  to  iX,  making  C^jy  equal  to 
B'C,  and  join  A'  and  IT. 

The  triangle  ADC  and  the  triangle  ABC  having  the  side 
A  0  in  common ;  the  side  CD  equal  to  the  side  CB  by  con- 
struction ;  and  the  included  angle  A  CD  equal  to  the  included 
angle  ACB,  since  they  are  adjacent  angles  and  ACB  is  ^ 
right  angle  by  hypothesis,  are  equal,  by  Proposition  YI. 

In  like  manner  the  triangle  A!D^C'  may  be  proved  equal 
to  the  triangle  MB'C, 

The  triangles  BAD  and  B'A!D^  having  the  side  AB  equal 
to  the  side  A!B!  by  hypothesis;  the  side  BD  equal  to  the 
side  B'D'^  because  they  were  constructed  the  doubles  oi  BG 
and  B'C'^  which  were  equal  by  hypothesis;  and  the  side  AD 
equal  to  the  side  A!D\  since  they  have  been  proved  to  be 
equal  respectively  to  the  sides  AB  and  A!B' ;  are  equal  to 
each  other,  by  Proposition  IX.,  and  B  is  equal  to  B'. 

The  triangles  ABC  and  A!B'C'  have  now  been  proved  to 
have  two  sides  and  the  included  angle  of  the  one  respectively 
equal  to  two  sides  and  the  included  angle  of  the  other,  and 
are  equal,  by  Proposition  VI. 


BOOK   I.  25 

PROPOSITION  XI.— THEOREM. 

27.  If  two  angles  of  a  triangle  are  equals  the  sides  opposite  to 
them  are  equals  and  the  triangle  is  isosceles. 

Let  the  angles  BAG  and  BCA  of  the  triangle  ABC  be 
equal,  then  are  the  sides  AB  and  BG  equal. 

For,  if  AB  and  BG  are  not  equal,  one  must 
be  greater  than  the  other.  Suppose  AB  greater 
than  BG. 

Then  cut  oif  from  AB  a  part  AD  equal  to 
BG,  and  join  D  and   G.     Compare   now  the 
triangle  ADG  with  the  whole  triangle  ABG,  of  which  it  is  a 
part. 

The  two  triangles  have  the  side  A  C  in  common ;  the  side 
AD  equal  to  the  side  BG  hy  construction;  and  the  included 
angle  A  equal  to  the  included  angle  BGA  by  hypothesis. 
Therefore,  by  Proposition  YI.,  the  triangles  ADG  and  ABG 
are  equal,  which  is  impossible.  Consequently,  AB  could  not 
have  been  greater  than  BG. 

In  like  manner  we  can  prove  that  BG  cannot  be  greater 
than  AB. 

Therefore,  since  neither  can  be  greater  than  the  other,  AB 
and  BG  are  equal. 

EXERCISE. 

Theorem. — An  equiangular  triangle  is  also  equilateral. 


^  r  n    (h 


26  ELEMENTS  OF  GEOMETRY. 

PKOPOSITION  XII.— THEOEEM. 

28.  If  two  angles  of  a  triangle  are  unequal^  the  side  opposite 
the  greater  angle  is  greater  than  the  side  opposite  the  less  angle. 

In  the  triangle  ABC  let  the  angle  C  be  greater  than  the 
angle  B  ;  then  AB  is  greater  than  A  C. 

For,  suppose  the  line  CD  to  be  drawn,  cutting 
off  from  the  greater  angle  a  part  BCD  =  B. 
Then  BDC  is  an  isosceles  triangle,  by  Propo- 
sition XI.,  and  DC  =  DB.  But  in  the  triangle 
ADC  we  have  ^D  +  DC  >  AC,  by  Axiom  I. ;  or, 
putting  DB  for  its  equal  DC,AD  +  DB:>AC;ovAB:>Aa 

PROPOSITION  XIII.— THEOREM. 

29.  If  two  sides  of  a  triangle  are  unequal,  the  angle  opposite 
the  greater  side  is  greater  than  the  angle  opposite  the  less  side. 

In  the  triangle  ABC  let  the  side  AB  be  greater  than  the 
side  BC;  then  will  the  angle  C  be  greater  than 
the  angle  A. 

For,  if  C  is  not  greater  than  J.,  it  must  be 
equal  to  A  or  less  than  A. 

C  cannot  be  equal  to  A,  for  in  that  case  AB     ^  ^ 

and  BC  would  be  equal,  by  Proposition  XI. 

C  cannot  be  less  than  J.,  for  in  that  case  AB  would  be 
less  than  BC,  by  Proposition  XII. 

Therefore  C  is  greater  than  A. 


BOOK  I.  27 

PROPOSITION  XIV.— THEOREM. 

30.  If  two  triangles  have  two  sides  of  the  one  respectively 
equal  to  two  sides  of  the  other,  and  the  included  angles  unequal, 
the  triangle  which  has  the  greater  included  angle  has  the  greater 
third  side. 

Let  ABC  and  ABB  be  the  two  triangles  in  which  the  sides 
AB,   AC  are  respectively  equal  to  the 
sides  AB,  AD,  but  the   included  angle 
BAC  is  greater  than  the  included  angle 
BAD;  then  ^Ois  greater  than  BD. 

For,  suppose  the  line  AE  to  be  drawn, 

bisecting  the  angle  CAD  and  meeting  BC 

in  E;  join  DE.     The  triangles  AED  and  AEC  are  equal,  by 

Proposition  YI.,  and  therefore  ED  =  EC.     But  in  the  triangle 

BDE  we  have 

BE  +  ED:>  BD,  by  Axiom  I., 

and  substituting  EC  for  its  equal  ED, 

BE+  EC>  BD,  or  BC  >  BD. 

PROPOSITION  XV.— THEOREM. 

31.  If  two  triangles  have  two  sides  of  the  one  respectively 
equal  to  two  sides  of  the  other,  and  the  third  sides  unequal,  the 
triangle  which  has  the  greater  third  side  has  the  greater  included 
angle. 

In  the  triangles  ABC  and  DEF  let  AB  =  DE,  AC  =  DF, 
and  let  5(7  be  greater  than  EF ; 
then  will  the   angle  A  be  greater 
than  the  angle  D. 

For,  if  A  were  equal  to  D,  BC 
would  be  equal  to  EF,  by  Proposi- 
tion YI. ;  and  if  A  were  less  than 
D,  BC  would  be  less  than  EF,  by  Proposition  XIY. 


28  ELEMENTS  OF  GEOMETRY. 

PEOPOSITION  XVI.— THEOEEM. 

32.  From  a  given  pointy  without  a  straight  line^  one  perpen- 
dicular  can  be  drawn  to  the  line^  and  but  one. 

Let  AB  be  the  given  line  and  P  the  given  point. 
Take  a  second  line  DJS,  and  at  some  point  F  of  DE  let 
a  perpendicular  be  erected  (Proposition  I.).     Superpose  this 

second  figure  upon  the 

p  ^ 

first,    placing    the    line 

DF  upon  the  line  AB, 
and  then  move  the  fig- 
ure along,  keeping  DF     ^  ^     J  J  J 
always    in    coincidence 

with  AB,  until  the  perpendicular  FGr  passes  through  P;  we 
shall  then  have  a  perpendicular  to  AB  drawn  through  P. 
Let  PC  in  the  figure  below  be  this  perpendicular. 

No  other  perpendicular  from  P  can  be  drawn  to  the  line  AB. 
For,  suppose  that  a  second  perpendicular  FD  could  be  drawn. 

Extend  FG  to  P',  making  GF'  equal  to  PC,  and  join  J> 
andP'. 

The  two  triangles  FGF  and  F'GF  have 
the  side  FG  equal  to  the  side  F'G  by  con- 
struction; the  side  GF  common;  and  the 
included  angle  FGF  equal  to  the  included 
angle  F'GD,  by  Proposition  III.  There- 
fore, by  Proposition  YI.,  the  triangles  are 
equal,  and  the  angle  FDG  is  equal  to  the 
angle  F'FG.  But  FDG  is  a  right  angle  by  hypothesis; 
therefore  F'DG  must  be  a  right  angle,  and  FD  and  DF 
must  lie  in  the  same  straight  line,  by  Proposition  lY. ;  and 
we  have  two  straight  lines  drawn  between  P  and  P',  which, 
by  Postulate  I.,  is  impossible. 


BOOK   I.  29 

Since  this  impossible  result  follows  necessarily  from  the 
assumption  that  a  second  perpendicular  can  be  drawn  from 
P  to  AB,  that  assumption  must  be  false. 

EXERCISES. 

1.  Theorem. — A  perpendicular  let  fall  from  the  vertex  of  an 
isosceles  triangle  upon  the  base  bisects  the  base  and  bisects  the 
vertical  angle. 

2.  Theorem. — Two  right  triangles  are  equal  when  they  have 
the  hypotenuse  and  an  adjacent  angle  of  the  one  respectively 
equal  to  the  hypotenuse  and  an  adjacent  angle  of  the  other. 

Suggestion.  Superpose  the  second  triangle  upon  the  first, 
making  the  given  equal  angles  coincide. 

PROPOSITION  XVII.— THEOREM. 

33.  The  perpendicular  is  the  shortest  line  that  can  be  drawn 
from  a  point  to  a  straight  line. 

Let  PC  be  the  perpendicular  and  PD 
any  oblique  line  from  the  point  ^P  to  the 
line  AB.     Then  PC  is  shorter  than  PD. 

Extend  PC  to  P',  making  CP'  equal  to 
PC,  and  join  D  and  P'. 

The  triangles  PCD  and  P'CD  are  equal, 
by  Proposition  YI.     Therefore  P'B  =  PD. 

PP'  <^PP  +  DP',  by  Axiom  I. 

Therefore  PC,  the  half  of  PP',  is  less  than  PD,  the  half 
of  PJ)P\ 

EXERCISES. 

1.  Theorem. —  Two  oblique  lines  drawn  from  a -point  to  a  line, 
and  meeting  the  line  at  equal  distances  from  the  foot  of  the  per- 
pendicular from  the  given  point,  are  equal. 

2.  Theorem. — Two  equal  oblique  lines  drawn  from  a  point  to 
a  line  meet  it  at  equal  distances  from  the  foot  of  the  perpen- 
dicular. 

3* 


30 


ELEMENTS   OF   GEOMETEY. 


PROPOSITION  XVIII.— THEOREM. 

34.  If  a  perpendicular  is  erected  at  the  middle  of  a  straight 
linCj  then, 

1st.  Uvery  point  in  the  perpendicular  is  equally  distant  from 
the  extremities  of  the  line  ; 

2d.  Every  point  without  the  perpendicular  is  unequally  distant 
from  the  extremities  of  the  line. 

Let  AB  be  a  finite  straight  line  and 
CD  a  perpendicular  at  its  middle  point. 

1st.  Then  is  any  point  P  on  CD  equi- 
distant from  A  and  B.  For,  join  P  and 
A  and  P  and  J5. 

The  triangles  FCA  and  FOB  are  equal, 
by  Proposition  YI. ;  therefore  FA  and  FB  are  equal. 

2d.  Any  point  Q  without  the  perpen- 
dicular is  unequally  distant  from  A  and 
B.    For,  Q  being  on  one  side  or  the  other 
of  the  perpendicular,  one  of  the  lines  QA, 
QB  must   cut  the  perpendicular;   let  it 
be  QA  and  let  it  cut  in  P;  join  FB.     The  straight  line  QB 
is  less  than  the  broken  line   QFB^  by  Axiom  I. ;  that  is, 
C^  <  §P  +  FB.     But  FB  =  FA  ;  therefore  QB  <:  QF  + 
FA,  or  QB  <  QA. 

35.  Definition.  A  geometric  locus  is  the  geometric  figure 
containing  all  the  points  which  possess  a  common  property, 
and  no  others. 

In  this  definition,  points  are  understood  to  have  a  common 
property  when  they  satisfy  the  same  geometrical  conditions. 

Thus,  since  all  the  points  in  the  perpendicular  erected  at 
the  middle  of  a  line  possess  the  common  property  of  being 
equally  distant  from  the  extremities  of  the  line  (that  is, 


BOOK  I.  31 

satisfy  the  condition  that  they  shall  be  equally  distant  from 
those  extremities),  and  no  other  points  possess  this  property^ 
the  perpendicular  is  the  locus  of  these  points;  so  that  the 
preceding  proposition  is  fully  covered  by  the  following  brief 
statement : 

The  perpendicular  erected  at  the  middle  of  a  straight  line  is 
the  locus  of  the  points  which  are  equally  distant  from  the  extrem- 
ities of  that  line, 

PROPOSITION  XIX.— THEOREM. 

36.  Uvery  point  in  the  bisector  of  an  angle  is  equally  distant 
from  the  sides  of  the  angle  ;  and  every  point  not  in  the  bisector, 
but  within  the  angle,  is  unequally  distant  from  the  sides  of  the 
angle ;  that  is,  the  bisector  of  an  angle  is  the  locus  of  the  points 
within  the  angle  and  equally  distant  from  its  sides. 

1st.  Let  AD  be  the  bisector  of  the  angle  JBAC,  P  any 
point  in  it,  and  PE,  PF,  the  perpendicular 
distances  of  P  from  AB  and  AC ;   then 
PE  =  PF. 

For,  the  right  triangles  APE,  APF, 
having  the  angles  PAE  and  PAF  equal, 
and  AP  common,  are  equal  (32,  Exercise 
2)  ;  therefore  PE  =  PF. 

2d.  Let  Q  be  any  point  not  in  the  bisector,  but  within 
the  angle;  then  the  perpendicular  distances  QE  and  QH 
are  unequal. 

For,  suppose  that  one  of  these  distances,  as  QE,  cuts  the 
bisector  in  some  point  P;  from  P  let  PF  be  drawn  perpen- 
dicular to  AC,  and  join  QF.  We  have  QJff  <C  QF ;  also^ 
§P  <  §P  +  PF,  or  QF<iQP  +  PE,  or  QF  <  QE  ^  there- 
fore qe:<  QE. 


32 


ELEMENTS   OF   GEOMETRY. 


When  the  angle  BAG  is  obtuse,  the  point  Q,  not  in  the 
bisector,  may  be  so  situated  that  the  perpendicular  on  one 
of  the  sides,  as  AB,  will  fall  at  the 
vertex  A ;  the  perpendicular  QS  is 
then  less  than  the  oblique  line  QA. 
Or,  a  point  §'  may  be  so  situated 
that  the  perpendicular  §'^',  let  fall 
on  one  of  the  sides,  as  AB,  will  meet 

that  side  produced  through  the  vertex  A  ;  this  perpendicular 
must  cut  the  side  AC  in.  some  point,  K^  and  we  then  have 

EXEBCISE. 

Theorem, — The  locus  of  the  points  equally  distant  from  two 
intersecting  straight  lines  is  the  pair  of  lines  which  bisect  all 
the  angles  formed  by  the  given  lines. 

(v.  19,  Exercise  2.) 

37.  Definition.  A  broken  line,  as  ABCBE^  is  called  convex 
when  no  one  of  its  component  straight 

lines,   if  produced,   can  enter  the   space  ^ 

enclosed    by   the    broken    line,    and   the 
straight  line  joining  its  extremities. 

PROPOSITION  XX.— THEOREM. 

38.  A  convex  broken  line  is  less  than  any  other  line  which 
envelops  it  and  has  the  same  extremities. 

Let  the  convex  broken  line  AFQE  have  the  same  extrem- 
ities A,  E^  as  the  line  ABODE,  and  be 
enveloped  by  it;  that  is,  wholly  in- 
cluded within  the  space  bounded  by 
ABCDE  and  the  straight  line  AE. 
Then  AFGE  <  ABCDE. 

For,  produce  AF  and  FG  to  meet  the  enveloping  line  in 


Ch 


BOOK   I.  33 

H  and  K.  Imagine  ABODE  to  be  the  path  of  a  point 
moving  from  A  to  E.  If  the  straight  line  AH  be  substituted 
for  ABCH,  the  path  AHDE  will  be  shorter  than  the  path 
ABODE,  the  portion  HDE  being  common  to  both.  If, 
further,  the  straight  line  FK  be  substituted  for  FHDK,  the 
path  AFKE  will  be  a  still  shorter  path  from  A  to  E.  And 
if,  finally,  GE  be  substituted  for  GKE,  AFGE  will  be  a  still 
shorter  path.  Therefore  AFGE  is  less  than  any  enveloping 
line. 

39.  Scholium.  The  preceding  demonstration  applies  when 
the  enveloping  line  is  a  curve,  or  any  species  of  line  whatever. 

/     PROPOSITION  XXI.— THEOREM. 

40.  If  two  oblique  lines  drawn  from  a  point  to  a  line  meet  the 
line  at  unequal  distances  from  the  foot  of  the  perpendicular,  the 
more  remote  is  the  greater. 

Ist.  If  the  lines  lie  on  the  same  side  of  the  perpendicular. 

Let  PO  be  the  perpendicular  and  FD  and  FE  the  two 
oblique  lines,  EO  being  greater  than 
DO;  then  is  FE  greater  than  FD. 

For,  produce  FO  to  P',  making  OP' 
equal  to  FO,  and  join  P'  with  D  and 
with  E. 

The  triangles  FDO  and  F'DO  and 
the  triangles  FEO  and  F^EO  are  equal, 
by  Proposition  YI.  ILence  FD  =  F'D, 
and  FE  =  F'E. 

FDF'  is  less  than  FEF\  by  Proposition  XX. ;  therefore 
FE,  the  half  of  FEF\  is  greater  than  FD,  the  half  of 
FDF\ 


34 


ELEMENTS  OF  GEOMETEY. 


2d.  If  the  lines  lie  on  opposite  sides  of  the  perpendicular. 

Let  PC  be  the  perpendicular  and  FD  and  PE  the  oblique 
lines,  EG  being  greater  than  CD. 

Lay  off  CD'  equal  to  CD,  and 
join  P  and  D'.  Then  the  triangles 
PDC  and  PD'C  are  equal,  by  Prop- 
osition YI.,  and  PD' =:  PD. 

But  PD'  is  less  than  PE,  by  the 
proof  given  above.    Hence  its  equal  PD  is  less  than  PE, 


PROPOSITION  XXII.— THEOREM. 

41.  Two  straight  lines  perpendicular  to  the  same  straight  line 
are  parallel. 

Let  AB  and  CD  be  two  lines  per- 
pendicular  to   the   same   line  EF;     

then  are  they  parallel.     For,  if  AB 

and  CD  are  not  parallel,  they  must     

meet  if  produced ;  but  this  is  impos- 
sible, for  in  that  case  we  should 
have  two  perpendiculars  from  their 

point  of  meeting  to  the  same  straight  line  EF,  which  is 
contrary  to  Proposition  XYI. 


PROPOSITION  XXIII.— THEOREM. 

42.  Through  a  given  point  one  line,  and  only  one,  can  be 
drawn  parallel  to  a  given  line. 

Let  A  be  the  given  point  and  — 

BC  the  given  line. 

From  A  draw  AD  perpendicular    ^ 

to  BC,  and  through  A  draw  AE 

perpendicular  to  AD.    AE  and  DC,  being  perpendicular  to 


BOOK   I.  35 

the  same  line  AI),  are  parallel,  by  Proposition  XXII.  No 
other  line  can  be  drawn  through  A  parallel  to  BC^  for,  by 
Axiom  II.,  it  would  have  the  same  direction  as  AE^  and 
therefore,  by  Postulate  II.,  it  would  coincide  with  AE, 


EXERCISES. 

1.  Theorem. — Lines  having  the  same  direction  are  parallel. 
Suggestion.  Suppose  them  to  meet ;  v.  Postulate  II. 

2.  Theorem. — Lines  parallel  to  the  same  line  are  parallel  to 
each  other. 

43.  Definitions.  When  two  straight  lines  AB,  CD,  are  cut 
by  a  third  EF,  the  eight  angles  formed 
at    their    points    of   intersection    are 
named  as  follows  : 

The  four  angles,  1,  2,  3,  4,  without 
the  two  lines,  are  called  exterior  angles. 

The  four  angles,  5,  6,  7,  8,  within 
the  two  lines,  are  called  interior  angles. 

Two  exterior  angles  on  opposite  sides  of  the  secant  line 
and  not  adjacent — as  1,  3 — or  2, 4 — are  called  alternate-exterior 
angles. 

Two  interior  angles  on  opposite  sides  of  the  secant  line 
and  not  adjacent — as  5,  7 — or  6,  8 — are  called  alternate-interior 
angles. 

Two  angles  similarly  situated  with  respect  both  to  the 
secant  and  to  the  line  intersected  by  it  are  caUed  correspond- 
ing angles ;  as  1,  5 — 2,  6 — 3,  7 — 4,  8. 


36  ELEMENTS   OF   GEOMETRY. 

PROPOSITION  XXIV.— THEOREM. 

44.  When  two  straight  lines  are  cut  by  a  third,  if  the  alter- 
nate-interior  angles  are  equal,  the  two  straight  lines  are  parallel. 

Let  the  line  AB  cut  the  lines  CD  and  UF,  making  the 
alternate-interior  angles  CAB  and  ABF  equal ;  then  are  CD 
and  FF  parallel. 

Through  Gr,  the  middle 
point  of  AB,  draw  GS  per- 
pendicular to  CD,  and  cut- 
ting FF  in  I. 

Then  the  triangles  AGH 
and  BGI,  having  the  side  AG  equal  to  the  side  GB  by- 
construction,  the  angle  GAH  equal  to  the  angle  GBI^y 
hypothesis,  and  the  angle  AGS  equal  to  the  angle  IGB  by- 
Proposition  Y.,  are  equal,  by  Proposition  YII.  Therefore  the 
angle  GIB  is  equal  to  the  angle  GHA.  But  GHA  is  a  right 
angle  by  construction ;  hence  GIB  is  a  right  angle,  and  CD 
and  FF  are  perpendicular  to  the  same  line  HI,  and  are 
therefore  parallel,  by  Proposition  XXII. 

If  FBA  and  BAD  are  the  given  equal  alternate  angles, 
their  supplements  CAB  and  ABF  are  equal,  and  the  proof 
just  given  is  valid. 

If  the  given  alternate-interior  angles  are  right  angles,  the 
lines  are  parallel,  by  Proposition  XXII. 

45.  Corollary  I.  When  two  straight  lines  are  cut  by  a  third, 
if  a  pair  of  corresponding  angles  are  equal,  the  lines  are  parallel. 

Suggestion.  Show  that  in  that  case  a  pair  of  alternate- 
interior  angles  are  equal. 

46.  Corollary  II.  When  two  straight  lines  are  cut  by  a 
third,  if  the  sum  of  two  interior  angles  on  the  same  side  of  the 
secant  line  is  equal  to  two  right  angles,  the  two  lines  are  parallel. 


BOOK   I.  37 

Suggestion.  Show  that  a  pair  of  alternate-interior  angles 
are  equal. 

PEOPOSITION  XXV.— THEOKEM. 

47.  If  two  parallel  lines  are  cut  by  a  third  straight  line,  the 
alternate-interior  angles  are  equal. 

Let  the  parallel  lines  CI>  and  EF  be  cut  by  the  line  AB ; 
then  will  the  angles  CAB  and  ABF  be  equal. 
For,  if  they  are  not  equal, 


draw  through  A  sl  line  AG, 
making  the  angles  GAB  and 
ABF  equal ;  then,  by  Propo- 
sition XXIV.,  GA  and  EF 
are   parallel,   and  we  have 

two  parallels  to  the  same  line  EF  drawn  through  the  same 
point  A,  which  is  contrary  to  Proposition  XXIII.,  and  there- 
fore impossible.    Hence  the  angles  CAB  and  ABF  are  equal. 

48.  Corollary  I.  If  two  parallel  lines  are  cut  by  a  third 
straight  line,  any  two  corresponding  angles  are  equal. 

49.  Corollary  II.  If  two  parallel  lines  are  cut  by  a  third 
straight  line,  the  sum  of  the  two  interior  angles  on  the  same  side 
of  the  secant  line  is  equal  to  two  right  angles. 

EXERCISE. 

Theorem. — A  line  perpendicular  to  one  of  two  parallel  lines  is 
perpendicular  to  the  other. 


/f^^  Off  rn^.'^^^ 


/^ .        ay 


J^IFOUll^ 


38 


ELEMENTS  OF  GEOMETRY. 


PK0P08ITI0N  XXVL— THEOREM. 


50.  The  sum  of  the  three  angles  of  any  triangle  is  equal  to 
two  right  angles. 

Let  ABC  he  any  triangle ;  then  the  sum  of  its  three  angles 
is  equal  to  two  right  angles. 

Produce  BGj  and  through  G  draw  C£J  parallel  to  BA» 
Since  the  line  AG  meets  the  parallel 

A 

lines  AB  and  EG,  the  alternate-interior 
angles  AGJ3  and  BAG  are  equal,  by 
Proposition  XXV. 

Since  the  line  BD  cuts  the  parallel 
lines  AB  and  EG,  the  corresponding  angles  EGD  and  ABG 
are  equal,  by  Proposition  XXY.,  Corollary  I. 

Therefore  the  sum  of  the  angles  of  the  triangle  is  equal 
to  BGA-\-  AGE  +  EGD,  which  is  two  right  angles,  by  Propo- 
sition III.,  Corollary  I. 

51.  C0R01.LARY.  If  a  side  of  a  triangle  is  eoctended,  the  exterior 
angle  is  equal  to  the  sum  of  the  two  interior  opposite  angles. 


EXERCISE. 


Theorem. — If  the  sides 
of  an  angle  are  respectively 
perpendicular  to  the  sides 
of  a  second  angle,  the  an- 
gles are  equal,  or  supple- 
mentary. 


BOOK  I.  39 

POLYGONS. 

52.  Definitions.  A  polygon  is  a  portion  of  a  plane  bounded 
by  straight  lines ;  as  ABODE.     The  bounding 

lines  are  the  sides  ;  their  sum  is  the  perimeter 
of  the  polygon.  The  angles  which  the  adja- 
cent sides  make  with  each  other  are  the  angles 
of  the  polygon;  and  the  vertices  of  these 
angles  are  called  the  vertices  of  the  polygon. 

Any  line  joining  two  vertices  not  consecutive  is  called  a 
diagonal;  as  AG. 

53.  Definitions.  Polygons  are  classed  according  to  the  num- 
ber of  their  sides : 

A  triangle  is  a  polygon  of  three  sides. 

A  quadrilateral  is  a  polygon  of  four  sides. 

A  pentagon  has  five  sides ;  a  hexagon,  six ;  a  heptagon,  seven ; 
an  octagon,  eight ;  an  enneagon,  nine  j  a  decagon^  ten ;  a  dodec- 
agon, twelve ;  etc. 

An  equilateral  polygon  is  one  all  of  whose  sides  are  equal ; 
an  equiangular  polygon,  one  all  of  whose  angles  are  equal. 

54.  Definition.  A  convex  polygon  is  one  no  side  of  which, 
when  produced,  can  enter  within  the  space  enclosed  by  the 
perimeter;  as  ABODE  in  (52).  Each  of  the  angles  of  such 
a  polygon  is  less  than  two  right  angles. 

It  is  also  evident  from  the  definition  that  the  perimeter 
of  a  convex  polygon  cannot  be  intersected  by  a  straight  line 
in  more  than  two  points. 

A  concave  polygon  is  one  of  which  two 
or  more  sides,  when  produced,  will  enter 
the  space  enclosed  by  the  perimeter;  as 
MNOPQ,  of  which  OP  and  QP,  when  ^ 
produced,  will  enter  within  the  polygon. 
The  angle  OPQ,  formed  by  two  adjacent  re-entrant  sides, 


40 


ELEMENTS  OF   GEOMETRY. 


is  called  a  re-entrant  angle ;  and  hence  a  concave  polygon  is 
^OTCLQiimQ^  cdXlQ^  2i  re-entrant  polygon. 

All  the  polygons  hereafter  considered  will  be  understood 
to  be  convex. 


PROPOSITION  XXVII.— THEOREM. 

55,  The  sum  of  all  the  angles  of  any  polygon  is  equal  to  twice 
as  many  right  angles,  less  four,  as  the  figure  has  sides. 

Join  any  point  0  within  the  polygon  to  each  of  the  ver- 
tices, thus  dividing  the  polygon  into  as  many  triangles  as  it 
has  sides. 

The  sum  of  the  angles  of  these  tri- 
angles will,  by  Proposition  XXYI., 
be  twice  as  many  right  angles  as  the 
figure  has  sides.  But  the  angles  of 
the  triangles  form  the  angles  of  the 
polygon  plus  the  angles  at  0,  which 
are  equal  to  four  right  angles,  by 
Proposition  III.,  Corollary  II. 


EXERCISE. 


1.  Theorem. — If  each  side  of  a  polygon  is  extended,  the  sum 
of  the  exterior  angles  is  four  right  angles. 


Suggestion.  The  sum  of  all  the  an- 
gles, exterior  and  interior,  is  obvi- 
ously twice  as  many  right  angles  as 
the  figure  has  sides. 


BOOK  I.  41 

QUADRILATERALS. 

56.  Definitions.  Quadrilaterals  are  divided  into  classes,  as 
follows : 

Ist.  The  trapezium  (J.),  which  has  no  two 
of  its  sides  parallel. 

2d.    The  trapezoid  (jB),  which  has  two 
sides  parallel.     The  parallel  sides  are  called 


the  hases^  and  the  perpendicular  distance      /^       ^    \  \ 
between  them  the  altitude  of  the  trapezoid. 

3d.  The  parallelogram  ( (7),  which  is  bounded     v r\ 

by  two  pairs  of  parallel  sides.  \ j   \ 

The   side  upon  which   a  parallelogram  is 
supposed  to  stand  and  the  opposite  side  are  called  its  lower 
and  upper  bases.    The  perpendicular  distance  between  the 
bases  is  the  altitude. 

57.  Definitions.  Parallelograms  are  divided  into  species,  as 
follows : 

1st.  The    rhomboid  (a),   whose   adjacent      's* 

sides  are  not  equal  and  whose  angles  are         \ 
not  right  angles. 


2d.  The  rhombus^  or  lozenge  (&),  whose  sides 
are  all  equal. 

3d.  The  rectangle  (c),  whose  angles  are  all 
equal,  and  therefore  right  angles. 


4th.  The  square  (d),  whose  sides  are  all  equal  and 
whose  angles  are  all  equal. 

The  square  is  at  once  a  rhombus  and  a  rectangle. 


42  ELEMENTS   OF   GEOMETRY. 

PROPOSITION  XXVIII.— THEOREM. 

58.  Two  parallelograms  are  equal  when  two  adjacent  sides 
and  the  included  angle  of  the  one  are  equal  to  two  adjacent  sides 
and  the  included  angle  of  the  other. 

AD  ==  A'jy,  and  the  angle  BAD      I j   ^/ j 

=  ^'J.'i/;  then  these  parallelo-     ^  ^     ^'  ^ 

grams  are  equal. 

For  they  may  evidently  be  applied  the  one  to  the  other,  so 
as  to  coincide  throughout,     (v.  Proposition  XXIII.) 

59.  Corollary.  Two  rectangles 
are  equal  when  they  have  equal  bases 
and  equal  altitudes. 

PROPOSITION  XXIX.— THEOREM. 

60.  The  opposite  sides  of  a  parallelogram  are  equal  and  the 
opposite  angles  are  equal. 

Suggestion.  Draw  a  diagonal  AC.    ACB  J^  ^ 

and  CAD  are  equal,  by  Proposition  XXY. 

CAB  and  AGD  are  equal,  by  Proposition 
XXY. 

Hence  the  triangles  ABC  and  ADC  are  equal,  by  Propo- 
sition YII. 

EXERCISES. 

1.  Theorem. — If  one  angle  of  a  parallelogram  is  a  right  angle, 
all  the  angles  are  right  angles,  and  the  figure  is  a  rectangle. 

2.  Theorem. — If  two  angles 
have  the  sides  of  one  respectively 
parallel  to  the  sides  of  the  other, 
they  are  equal,  or  supplementary. 

3.  Theorem.  —  Two  parallel 
lines  are  everywhere  equidistant. 


v. 


BOOK   I.  43 

PROPOSITION  XXX.— THEOREM. 

61.  If  two  opposite  sides  of  a  quadrilateral  are  equal  and 
parallel,  the  figure  is  a  parallelogram. 

Suggestion.  Let  AD  be  equal  and  parallel  i^ ^ 

to  BC.    Draw  a  diagonal  AC.  /     \ 

The  triangles  ABC  and  ABO  are  equal,     £ 
by  Proposition  YI.     Therefore  the  angles 
BAG  and  ACD  are  equal,  and  AB  and  CD  are  parallel,  by 
Proposition  XXIY. 

PROPOSITION  XXXI.— THEOREM. 

62.  If  the  opposite  sides  of  a  quadrilateral  are  equals  the 
figure  is  a  parallelogram. 

Suggestion.  Draw  a  diagonal,  and  prove  the  two  triangles 
equal. 

PROPOSITION  XXXII.— THEOREM. 

63.  The  diagonals  of  a  parallelogram  bisect  each  other. 

A  D 

Suggestion.  The  triangles  AED  and  BEC 
are  equal,  by  Proposition  YII. 


EXERCISES. 

1.  Theorem. — The  diagonals  of  a  rectangle  are  equal. 

2.  Theorem. —  The  diagonals  of  a  rhombus  are  perpendicular 
to  each  other. 

3.  Theorem. — If  the  diagonals  of  a  quadrilateral  bisect  each 
other,  the  figure  is  a  parallelogram. 

4.  Theorem. — If  the  diagonals  of  a  parallelogram  are  equal, 
the  figure  is  a  rectangle. 

5.  Theorem. — If  the  diagonals  of  a  parallelogram  are  perpen- 
dicular to  each  other,  the  figure  is  a  rhombus. 


44 


ELEMENTS   OF   GEOMETRY. 


ARRANGEMENT  OF  WRITTEN  EXERCISES. 


64.  In  writing  out  a  demonstration,  brevity  of  statement 
and  clearness  of  arrangement  should  be  carefully  studied, 
and  symbols  and  abbreviations  may  bo  used  with  profit. 
The  following  list  is  recommended : 


SYMBOLS  AND 

ABBREVIATIONS. 

.  • .    therefore. 

Bef. 

definition. 

=     equal  to. 

Fost 

postulate. 

O     equivalent  to. 

Ax. 

axiom. 

>    greater  than. 

Prop. 

proposition. 

<;     less  than. 

Cor. 

corollary. 

II      parallel  to. 

Hyp. 

hypothesis. 

J_     perpendicular  to. 

Cons. 

construction. 

/_     angle. 

Adj. 

adjacent. 

^     angles. 

Inc. 

included. 

rt.  /_  right  angle. 

Alt.-int. 

alternate-interior. 

l\     triangle. 

Sup. 

supplementary. 

l^     triangles. 

Comp. 

complementary. 

rt.  I\  right  triangle. 

Q.E.B, 

,  quod  erat  demonstran- 

/     /  parallelogram. 

dum  (=  which  was 

/  *  /  parallelograms. 

to  be  proved). 

O     circle. 

0     circles. 

BOOK   I.  45 

65.  In  arranging  a  written  demonstration,  it  is  well  to 
begin  each  statement  on  a  separate  line,  giving  the  reason 
for  the  statement  at  the  end  of  the  line,  if  it  can  be  written 
briefly,  or  in  parenthesis  immediately  below  the  line,  if  it 
cannot  be  written  briefly.  The  following  examples  of  demon- 
strations prepared  as  written  exercises,  or  for  a  written 
examination,  will  serve  as  illustrations. 


(1)  PROPOSITION  XII.— THEOREM. 

If  two  angles  of  a  triangle  are  unequal^  the  side  opposite  the 
greater  angle  is  greater  than  the  side  opposite  the  less  angle. 


In  A  ABC,  let  it  be  given  that  Z_ACBy  /^B, 
we  are  to  prove  AB  >  J.  (7. 

Draw  CDj  cutting  off  from  /^ACB  si  part 
/  BCD  =  /_B. 
Then  in  /\  BCD  we  have 

/  BCD  =  l^B.  Cons. 

BD  =  CD,  Prop.  XL 

and  BD  +  DA  =  CD  ^  DA. 

But  AC<:CD-^DA.  Ax.  I. 

AC<iBD  ^DA, 
^^^  AC<:^AB,  Q.  E.  D. 


46  ELEMENTS  OF  GEOMETRY. 

(2)  PROPOSITION  XXIV.— THEOREM. 

When  two  straight  lines  are  cut  by  a  third,  if  the  alternate* 
interior  angles  are  equal,  the  two  straight  lines  are  parallel. 


Let  AB  cut  CD  and  EF  in  the  points  A  and  B,  making 
/^BAG=  /^ABF, 
we  are  to  prove  CD  \\  to  EF. 

Through  G,  the  middle  point  of  AB,  draw  HI  ]_  to  CD. 

Then  in  the  A  ^^^  and  BGI 
we  have 


and 

But 


Aa^BG, 

Cons. 

/_GAH=  /_  GBI, 

Hyp. 

/_AGH=  /_^BGI. 

Prop.  Y. 

/\AGII=:  ^BGI, 

Prop.  YII. 

/_  GIB  =  /_  GHA 

' 

(homologous  angles  of  = 

A). 

/  G^^isar^.  /. 

Cons. 

/_GIBiBSirt.  /_, 

Him  ]_to  EF. 

J?7is_|_to  CD. 

Cons. 

CDand  j^i^are||. 

Prop.  XXII. 

Q. 

E.D. 

and 
But 


If  the  given  equal  ^  are  ABE  and  BAD, 
we  have  /  ABE  =  /_  BAD,  Hyp. 

/  ABE  +  /  ABF  =2rt.  /^,      Prop.  III. 
/  BAD  -{-  Z,S^0=2rt.  ^.     Prop.  III. 
/^ABF=  /_BAC, 
and  the  proof  given  above  applies. 


BOOK   I.  47 

If  the  given  equal  alt.-int.  /^  are  rt.  ^  the  two  given  lines 
are    I    to  the  third  line  and  are  ||,  by  Proposition  XXII. 


(3)  PEOPOSITION  XXVI.    COEOLLARY. 

If  one  side  of  a  triangle  is  extended^  the  exterior  angle  is  equal 
to  the  sum  of  the  two  interior  opposite  angles. 


B 


In  the  /^  ABC  let  the  side  AC  he  extended, 
we  are  to  prove    ^  BCD  ^  /_A  -\-  /_B. 
We  have  the  sum  of  the  adj.  ^ 

BCD  -{-BGA  =  2rt.  ^.  Prop.  III. 

But        /_A  +  /^B+  /^BCA  =  2rt.  ^.    Prop.  XXYL 

/^BCD==  /_A-\-  /^B.  Q.  E.  B. 


(4)      EXERCISE  3,  PAGE  43.— THEOREM. 

If  the  diagonals  of  a  quadrilateral  bisect  each  other j  the  figure 
is  a  parallelogram. 


..--'.B**. 


In  the  quadrilateral  ABCD^  let  the  diagonals  AD  and  BG 
bisect  each  other. 
i  We  are  to  prove  ABCD  a  /    "], 


48  ELEMENTS   OF   GEOMETRY. 

In  the  ^  AEB  and  CED 
we  have  GE  =  EB,  Hyp. 

ED  =  AE,  Hyp. 

/  CED  =  /_  AEB.  Prop.  Y. 

A  ^^^  =  A  C'^A  I'rop.  YI. 

and  OD  =  A^ 

(homologous  sides  of  equal  ^^ 
and  /_EDC=  Z_EAB 

(homologous  /»  of  equal  /^). 
But  JE^DCand  jEAJ5are  alt.-int.  ^. 

CD  is  II  to  AB,  Prop.  XXIY. 

and  since  CD  =  AB,  Proved  above. 

^^CD  is  a  O-  ^^^P-  ^^^' 

Q.  E.  D. 


EXERCISES  ON  BOOK  I. 


k 


1.  The  straight  line  AJEJ  which  bisects  the  angle 
exterior  to  the  vertical  angle  of  an  isosceles  tri- 
angle ABCis  parallel  to  the  base  BC, 


2.  If  from  a  variable  point  in  the  base  of  an  isos- 

1/  celes  triangle  parallels  to  the  sides  are  drawn,  a 

parallelogram  is  formed  whose  perimeter  is  constant. 


3.  The  sum  of  the  four  lines  drawn  to  the  vertices  of  a  quadri- 
f^ lateral  from  any  point  except  the  intersection  of  the  diagonals,  is 

greater  than  the  sum  of  the  diagonals. 
^       4.  The  lines  drawn  from  the  extremities  of  the  base  of  an  isos- 
celes triangle  to  the  middle  points  of  the  opposite  sides  are  equal. 
J       6.  The  perpendiculars  from  the  extremities  of  the  base  of  an 
•^  isosceles  triangle  upon  the  opposite  sides  are  equal. 

6.  The  bisectors  of  the  base  angles  of  an  isosceles  triangle  are 
equal. 

7.  A  perpendicular  let  fall  from  one  end  of  the  base  of  an  isos- 
celes triangle  upon  the  opposite  side  makes  with  the  base  an 
angle  equal  to  one-half  the  vertical  angle. 

8.  If  the  vertical  angle  of  an  isosceles  triangle  is  one-half  as 
great  as  an  angle  at  the  base,  a  line  bisecting  a  base  angle  will 
divide  the  given  triangle  into  two  isosceles  triangles. 

9.  If  two  isosceles  triangles  have  the  sides  of  one  equal  to  the 
sides  of  the  other,  and  the  base  of  one  double  the  altitude  of  the 

*/      other,  the  two  triangles  are  of  the  same  size. 

c      d  5  49 


60 


ELEMENTS  OP  GEOMETRY. 


10.  If  from  a  variable  point  P  in  the  base  of  an 
isosceles  triangle  ABC^  perpendiculars,  PM^  PN^ 
to  the  sides  are  drawn,  the  sum  of  PM  and  PN 
is  constant,  and  equal  to  the  perpendicular  from 
C  upon  AB. 

Suggestion,  The  triangles  PNC  and  PEC  are 
equal,  by  Proposition  VII. 


y 


11.  The  line  joining  the  feet  of  perpendiculars  let  fall  from  the 
extremities  of  the  base  of  an  isosceles  triangle  upon  the  opposite 
sides  is  parallel  to  the  base. 


12.  If  BE  bisects  the  angle  ^  of  a  triangle 
ABC^  and  CE  bisects  the  exterior  angle  ACD^ 
the  angle  E  is  equal  to  one-half  the  angle  A, 


13.  The  medial  line  to  any  side  of  a  triangle  is  less 
than  the  half  sum  of  the  other  two  sides. 

Definition.  A  line  joining  a  vertex  of  a  triangle 
with  the  middle  point  of  the  opposite  side  is  called 
a  medial  line. 


V 


14.  If  from  two  points,  A  and  -B,  on  the 
same  side  of  a  straight  line  JfiV,  straight 
lines,  APj  BPy  are  drawn  to  a  point  P  in 
that  line,  making  with  it  equal  angles  APM 
and  BPN^  the  sum  of  the  lines  AP  and  BP 
is  less  than  the  sum  of  any  other  two  lines, 
AQ  and  BQ^  drawn  from  A  and  B  to  any 
other  point  Q  in  MN. 


/   16.  If  the  medial  line  from  the  vertex  of  a  triangle  to  the  base 
^  is  equal  to  one-half  the  base,  the  vertical  angle  is  a  right  angle. 

^  16.  The  altitude  of  a  triangle  divides  the  vertical  angle  into 
two  parts,  whose  difference  is  equal  to  the  difference  of  the  base 
angles  of  the  triangle. 


BOOK  I. 


51 


17.  The  perpendicular  erected  at  the  middle  point 
^    of  one  side  of  a  triangle  meets  the  longer  of  the 
other  two  sides. 


18.  Lines  drawn  from  a  point  within 
a  triangle  to  the  extremities  of  the  base 
include  an  angle  greater  than  the  verti- 
cal angle  of  the  triangle,    (v.  Proposi- 


tion XXVI.,  Corollary.) 


19.  The  sum  of  the  angles  at  the  vertices  of 
a  five-pointed  star  (pentagram)  is  equal  to 
two  right  angles. 


20.  The  three  perpendiculars  erected  at  the 
middle  points  of  the  sides  of  a  triangle  meet 
in  the  same  point. 

Suggestion.  The  point  of  intersection  of 
EH  and  FK  is  equidistant  from  the  three 
vertices,  and  therefore  must  lie  on  DQ 
(Proposition  XVIII.). 

21.  The  three  bisectors  of  the  three  an- 
gles of  a  triangle  meet  in  the  same  point. 

Suggestion.  The  point  of  intersection  of 
BE  and  CF  is  equidistant  from  the  three 
sides,  and  therefore  must  lie  on  AD 
(Proposition  XIX.). 


22.  The  bisectors  of  two  external  angles  of  a  triangle  and  the 
bisector  of  the  remaining  internal  angle  meet  in  a  point. 


23.  If  from  the  diagonal  BD  of  a  square  ABCD, 
BE  is  cut  off  equal  to  BC,  and  EF  is  drawn  per- 
pendicular to  BDy  then  DE  =  EF=  FG, 


52 


ELEMENTS  OF  GEOMETRY. 


24.  In  a  trapezoid,  the  straight  line  joining  A 

the  middle  points  of  the  non-parallel  sides  is  / 

parallel  to  the  bases,  and  is  equal  to  one-half  y"^ 

their  sum.  L 

Suggestion.  Draw  HO  parallel  to  AB^  and  -B 
extend  ^Z).   DOF=  Offi^CProposition  VII.), 
and  EFHB  is  a  parallelogram,  by  Proposition  XXX. 


D    Q 


I 


H     0 


25.  If  the  sides  of  a  trapezoid  which  are 
"^    not  parallel  are  equal,  the  base  angles  are 
equal  and  the  diagonals  are  equal. 


A    E 


26.  If  through  the  four  vertices  of  a  quadrilateral  lines  are 
^  drawn  parallel  to  the  diagonals,  they  will  form  a  parallelogram 
twice  as  large  as  the  quadrilateral. 


C 


27.  The  three  perpendiculars  from 
^  the  vertices  of  a  triangle  to  the  op- 

posite  sides  meet  in  the  same  point. 
Suggestion.  Draw  through  the 
three  vertices  lines  parallel  to  the 
opposite  sides  of  the  triangle.  By 
the  aid  of  the  three  parallelograms 
ABCB',  ABA'C,  and  ACBG',  prove 
that  the  sides  of  A'B^C  are  bisected 
by  -4,  -B,  and  C,  See  now  Exercise  20. 

28.  If  a  straight  line  drawn  parallel  to  the  base 
^  of  a  triangle  bisects  one  of  the  sides,  it  also  bisects 

the  other  side ;  and  the  portion  of  it  intercepted 
between  the  two  sides  is  equal  to  one-half  the  base. 
Suggestion.  Draw  DF  parallel  to  A  C.    See  now 
Proposition  VII.  and  Proposition  XXIX. 

y   29.  The  straight  line  joining  the  middle  points  of  two  sides  of 
a  triangle  is  parallel  to  the  third  side.    {v.  Exercise  28.) 
* 
vT     30.  The  three  straight  lines  joining  the  middle  points  of  the 
sides  of  a  triangle  divide  the  triangle  into  four  equal  triangles. 

^      31.  In  any  right   triangle,   the    straight   line  ^ 

,    drawn  from  the  vertex  of  the  right  angle  to  the 
"^    middle  of  the  hypotenuse  is  equal  to  one-half 
the  hypotenuse,    (v.  Exercise  28.)  b 


BOOK   I. 


53 


/^  32.  The  straight  lines  joining  the  middle 
points  of  the  adjacent  sides  of  any  quadrilat- 
eral form  a  parallelogram  whose  perimeter  is 
equal  to  the  sum  of  the  diagonals  of  the  quad- 
rilateral (Exercise  29). 

33.  If  E  and  -Pare  the  middle  points  of  the 
/    opposite  sides,  AD,  BC,  of  a  parallelogram 
ABCD,  the  straight  lines  BE,  DF,  trisect  the 
diagonal  AC  (Exercise  28). 

/^  34.  The  four  bisectors  of  the  angles 
of  a  quadrilateral  form  a  second  quad- 
rilateral, the  opposite  angles  of  which 
are  supplementary. 

If  the  first  quadrilateral  is  a  paral- 
lelogram, the  second  is  a  rectangle. 
If  the  first  is  a  rectangle,  the  second 
is  a  square. 

36.  The  point  of  intersection  of  the  diagonals  of  a  parallelogram 
bisects  every  straight  line  drawn  through  it  and  terminated  by 
the  sides  of  the  parallelogram. 

36.  If  from  each  vertex  of  a  parallelogram 
the  same  given  distance  is  laid  ofi"  on  a  side 
of  the  parallelogram,  care  being  taken  that 
no  two  distances  are  laid  off  on  the  same 
side,  the  points  thus  obtained  will  be  the 
vertices  of  a  new  parallelogram. 

37.  If  from  two  opposite  vertices  of  a  par- 
allelogram equal  distances  are  laid  off  on 
the  sides  adjacent  to  those  vertices,  the 
points  thus  obtained  will  be  the  vertices 
of  a  parallelogram. 


B 

c 

0 

K 

:v 

1 

A            A' 
D 

0 

£■: 

■:--) 

7 

A    A' 


38.  The  three  medial  lines  of  a  triangle 
m.eet  in  the  same  point. 

Suggestion.  Let  O  be  the  point  of  inter- 
section of  AD  and  BE,  and  H  and  O  the 
middle  points  of  OB  and  OA.  Hence  prove 
OD  =  ^AD  and  OE  =  ^BE.  In  like  man- 
ner the  point  of  intersection  of  AD  and  CE  ^ 
can  be  shown  to  cut  off  one-third  of  AD. 

5* 


54 


ELEMENTS  OF  GEOMETRY. 


89.  The  intersection  of  the  straight 
lines  which  join  the  middle  points  of 
opposite  sides  of  any  quadrilateral  is 
the  middle  point  of  the  straight  line 
which  joins  the  middle  points  of  the 
diagonals  (Exercise  29). 


he 
he 


a«i 


b 


SYLLABUS  TO  BOOK  L 

POSTULATES,  AXIOMS,  AND  THEOKEMS. 


POSTULATE   I. 


Through  any  two  given  points  one  straight  line,  and  only  one,  can 

be  drawn. 

POSTULATE   II. 

Through  a  given  point  one  straight  line,  and  only  one,  can  be  drawn 

having  any  given  direction. 

AXIOM   I. 

A  straight  line  is  the  shortest  line  that  can  be  drawn  between  two 
points. 

AXIOM   IL 

Parallel  lines  have  the  same  direction. 

PROPOSITION  I. 

At  a  given  point  in  a  straight  line  one  perpendicular  to  the  line  can 
be  drawn,  and  but  one. 

Corollary.  Through  the  vertex  of  any  given  angle  one  straight  line 
can  be  drawn  bisecting  the  angle,  and  but  one. 

PROPOSITION   IL 

All  right  angles  are  equal. 

PROPOSITION  III.      ^ 

The  two  adjacent  angles  which  one  straight  line  makes  with  another 
are  together  equal  to  two  right  angles. 

Corollary  I.  The  sum  of  all  the  angles  having  a  common  vertex,  and 
formed  on  one  side  of  a  straight  line,  is  two  right  angles. 

Corollary  IL  The  sum  of  all  the  angles  that  can  be  formed  about  a 
point  in  a  plane  is  four  right  angles. 

PROPOSITION  IV. 

If  the  sum  of  two  adjacent  angles  is  two  right  angles,  their  exterior 
sides  are  in  the  same  straight  line. 

i 


ELEMENTS   OF   GEOMETRY. 


PKOPOSITION   V. 


If  two  straight  lines  intersect  each  other,  the  opposite  (or     .  *■ 
angles  are  equal. 

PKOPOSITION   VI. 

Two  triangles  are  equal  when  two  sides  and  the  included  ang^v  w;  ihtt 
one  are  respectively  equal  to  two  sides  and  the  included  angk  of  the 
other. 

PROPOSITION   VII 

Two  triangles  are  equal  when  a  side  and  the  two  adjacent  angles  «)f  the 
one  are  respectively  equal  to  a  side  and  the  two  adjacent  angh  "  tlin 
other. 

~-^.    PROPOSITION   VIII. 

In  an  isosceles  triangle  the  angles  opposite  the  equal  sides  are  equal. 
Corollary.  The  straight  line  bisecting  the  vertical  angle  of  an  isosceles 
triangle  bisects  the  base,  and  is  perpendicular  to  the  base. 

PROPOSITION   IX. 

Two  triangles  are  equal  when  the  three  sides  of  the  one  are  respectively 
equal  to  the  three  sides  of  the  other. 

PROPOSITION  X. 

Two  right  triangles  are  equal  when  they  have  the  hypotenus  :   •. 

side  of  the  one  respectively  equal  to  the  hypotenuse  and  a  sid( 
other. 

PROPOSITION   XI. 

If  two  angles  of  a  triangle  are  equal,  the  sides  opposite  to  tt 
equal,  and  the  triangle  is  isosceles. 

PROPOSITION   XII. 

If  two  angles  of  a  triangle  are  unequal,  the  side  opposite  the 
angle  is  greater  than  the  side  opposite  the  less  angle. 

PROPOSITION   XIII. 

If  two  sides  of  a  triangle  are  unequal,  the  angle  opposite  the 
side  is  greater  than  the  angle  opposite  the  less  side. 

PROPOSITION   XIV. 

If  two  triangles  have  two  sides  of  the  one  respectively  equal        '  wo 
sides  of  the  other,  and  the  included  angles  unequal,  the  triangle 
has  the  greater  included  angle  has  the  greater  third  side. 


71 


ELEMENTS   OF   GEOMETRY.  HI 

PROPOSITION    XV.  "^ 

If  two  triangles  have  two  sides  of  the  one  respectively  equal  to  two 
sides  of  the  other,  and  the  third  sides  unequal,  the  triangle  which  has  the 
greater  third  side  has  the  greater  included  angle. 

PROPOSITION    XVI. 

From  a  given  point,  without  a  straight  line,  one  perpendicular  can  be 
3ut  one. 

PROPOSITION   XVII. 

The  perpendicular  is  the  shortest  line  that  can  be  drawn  from  a  point 
to  a  straight  line. 

PROPOSITION   XVIII. 

If  a  perpendicular  is  erected  at  the  middle  of  a  straight  line,  then 
ever}^  point  on  the  perpendicular  is  equally  distant  from  the  extremities 
of  the  line ;  and  every  point  not  on  the  perpendicular  is  unequally 
distant  from  the  extremities  of  the  line. 

PROPOSITION   XIX. 

Every  point  in  the  bisector  of  an  angle  is  equally  distant  from  the 
sides  of  the  angle  ;  and  every  point  not  in  the  bisector,  but  within  the 
angle,  is  unequally  distant  from  the  sides  of  the  angle;  that  is,  the 
bisector  of  an  angle  is  the  locus  of  the  points  within  the  angle  and 
equally  distant  from  its  sides. 

PROPOSITION   XX. 

A  convex  broken  line  is  less  than  any  other  line  which  envelops  it 
and  has  the  same  extremities. 

PROPOSITION   XXI. 

If  two  oblique  lines  drawn  from  a  point  to  a  line  meet  the  line  at 
unequal  distances  from  the  foot  of  the  perpendicular,  the  more  remote 
is  the  greater. 

PROPOSITION   XXII. 

Two  straight  lines  perpendicular  to  the  same  straight  line  are  parallel. 

PROPOSITION   XXIII. 

Through  a  given  point  one  line,  and  only  one,  can  be  drawn  parallel 
to  a  given  line. 


f 


BOOK  11. 

THE   CIRCLE. 

1.  Definitions,  A  circle  is  a  portion  of  a  plane  bounded  t)y 
a  curve,  all  the  points  of  which  are  equally  distant  from  a 
point  within  it  called  the  centre. 

The  curve  which  bounds  the  circle  is 
called  its  circumference. 

Any  straight  line  drawn  from  the  cen- 
tre to  the  circumference  is  called  a 
radius. 

Any  straight  line  drawn  through  the 
centre  and  terminated  each  way  by  the  circumference  is 
called  a  diameter. 

In  the  figure,  0  is  the  centre,  and  the  curve  ABCEA  is 
the  circumference  of  the  circle;  the  circle  is  the  space  in- 
cluded within  the  circumference;  OA,  OBy  OC,  are  radii; 
-4.0(7  is  a  diameter. 

By  the  definition  of  a  circle,  all  its  radii  are  equal;  also 
all  its  diameters  are  equal,  each  being  double  the  radius. 

If  one  extremity,  0,  of  a  line  OA  is  fixed,  while  the  line 
revolves  in  a  plane,  the  other  extremity,  J.,  will  describe  a 
circumference,  whose  radii  are  all  equal  to  OA. 

2.  Definitions.  An  arc  of  a  circle  is  any  portion  of  its  cir- 
cumference ;  as  DEF. 

A  chord  is  any  straight  line  joining  two  points  of  the  cir- 
cumference; as  DF.  The  arc  DFF  is  said  to  be  subtended 
by  its  chord  DF. 

55 


ELEMENTS   OF   GEOMETRY.  ill 

PROPOSITION    XV.  "^ 

If  two  triangles  have  two  sides  of  the  one  respectively  equal  to  two 
sides  of  the  other,  and  the  third  sides  unequal,  the  triangle  which  has  the 
greater  third  side  has  the  greater  included  angle. 

PROPOSITION    XVI. 

From  a  given  point,  without  a  straight  line,  one  perpendicular  can  be 
drawn  to  the  line,  and  but  one. 

PROPOSITION    XVII. 

The  perpendicular  is  the  shortest  line  that  can  be  drawn  from  a  point 
to  a  straight  line. 

PROPOSITION   XVIII. 

If  a  perpendicular  is  erected  at  the  middle  of  a  straight  line,  then 
ever}'-  point  on  the  perpendicular  is  equally  distant  from  the  extremities 
of  the  line ;  and  every  point  not  on  the  perpendicular  is  unequally 
distant  from  the  extremities  of  the  line. 

PROPOSITION   XIX. 

Every  point  in  the  bisector  of  an  angle  is  equally  distant  from  the 
sides  of  the  angle  ;  and  every  point  not  in  the  bisector,  but  within  the 
angle,  is  unequally  distant  from  the  sides  of  the  angle;  that  is,  the 
bisector  of  an  angle  is  the  locus  of  the  points  within  the  angle  and 
equally  distant  from  its  sides. 

PROPOSITION   XX. 

A  convex  broken  line  is  less  than  any  other  line  which  envelops  it 
and  has  the  same  extremities. 

PROPOSITION   XXI. 

If  two  oblique  lines  drawn  from  a  point  to  a  line  meet  the  line  at 
unequal  distances  from  the  foot  of  the  perpendicular,  the  more  remote 
is  the  greater. 

PROPOSITION   XXII. 

Two  straight  lines  perpendicular  to  the  same  straight  line  are  parallel. 

PROPOSITION   XXIII. 

Through  a  given  point  one  line,  and  only  one,  can  be  drawn  parallel 
to  a  given  line. 


( 


BOOK  11. 

THE    CIRCLE. 

1.  Definitions.  A  circle  is  a  portion  of  a  plane  bounded  Dy 
a  curve,  all  the  points  of  which  are  equally  distant  from  a 
point  within  it  called  the  centre. 

The  curve  which  bounds  the  circle  is 
called  its  circumference. 

Any  straight  line  drawn  from  the  cen- 
tre to  the  circumference  is  called  a 
radius. 

Any  straight  line  drawn  through  the 
centre  and  terminated  each  way  by  the  circumference  is 
called  a  diameter. 

In  the  figure,  0  is  the  centre,  and  the  curve  ABCEA  is 
the  circumference  of  the  circle;  the  circle  is  the  space  in- 
cluded within  the  circumference;  OA^  OB,  OC,  are  radii; 
J.  0(7  is  a  diameter. 

By  the  definition  of  a  circle,  all  its  radii  are  equal;  also 
all  its  diameters  are  equal,  each  being  double  the  radius. 

If  one  extremity,  0,  of  a  line  OA  is  fixed,  while  the  line 
revolves  in  a  plane,  the  other  extremity.  A,  will  describe  a 
circumference,  whose  radii  are  all  equal  to  OA. 

2.  Definitions.  An  arc  of  a  circle  is  any  portion  of  its  cir- 
cumference ;  as  DEF. 

A  chord  is  any  straight  line  joining  two  points  of  the  cir- 
cumference;  as  DF.  The  arc  DEF  is  said  to  be  subtended 
by  its  chord  DF. 

55 


66  ELEMENTS   OF   GEOMETRY. 

Every  chord  subtends  two  arcs,  which  together  make  up 
the  whole  circumference.  Thus,  DF  subtends  both  the  arc 
DBF  and  the  arc  DGBAF.  When  an 
arc  and  its  chord  are  spoken  of,  the  arc 
less  than  a  semi-circumference,  as  DFF, 
is  always  understood,  unless  otherwise 
stated. 

A  segment  is  a  portion  of  the   circle 
included  between  an  arc  and  its  chord; 
thus,  by  the  segment  DFF  is  meant  the  space  included 
between  the  arc  BF  and  its  chord. 

A  sector  is  the  space  included  between  an  arc  and  the  two 
radii  drawn  to  its  extremities ;  as  A  OB. 

3.  From  the  definition  of  a  circle  it  follows  that  every 
point  within  the  circle  is  at  a  distance  from  the  centre  which 
is  less  than  the  radius ;  and  every  point  without  the  circle  is 
at  a  distance  from  the  centre  which  is  greater  than  the 
radius.  Hence  the  locus  of  all  the  points  in  a  plane  which 
are  at  a  given  distance  from  a  given  point  is  the  circumference 
of  a  circle  described  with  the  given  point  as  a  centre  and  with 
the  given  distance  as  a  radius. 

4.  Postulate.  A  circumference  may  be  described  with  any 
point  as  centre  and  any  distance  as  radius. 


PEOPOSITION  I.— THEOREM. 

5.  Two  circles  are  equal  when  the  radius  of  the  one  is  equal 
to  the  radius  of  the  other.  \ 

Let  the  second  circle  be  superposed  upon  the  first,  so  that 
its  centre  falls  upon  the  centre  of  the  first ;  then  will  the  two 
circumferences  coincide  throughout. 


BOOK   II.  57 

For,  if  any  point  of  either  circumference  falls  outside  of 
the  other  circle,  the  line  joining  it  with  the  common  centre 
must  cross  the  circumference  of  that  circle. 

The  whole  line  will  be  a  radius  of  one  circle,  the  portion 
of  it  within  the  other  circle  will  be  a  radius  of  that  other 
circle,  and  we  shall  have  two  unequal  radii,  which  is  con- 
trary to  our  hypothesis. 


PROPOSITION  II.— THEOREM. 

6.  Every  diameter  bisects  the  circle  and  its  circumference. 

Let  AMBN  be  a  circle  whose  centre  is 
0 ;  then  any  diameter  AOB  bisects  the 
circle  and  its  circumference. 

For,  if  the  figure  ANB  be  turned 
about  AB  as  an  axis  and  superposed 
upon  the  figure  AMB^  the  curve  ANB 
will  coincide  with  the  curve  AMB,  since 
all  the  points  of  both  are  equally  distant  from  the  centre. 
(v.  Proof  of  Proposition  I.)  The  two  figures  then  coincide 
throughout,  and  are  therefore  equal  in  all  respects.  There- 
fore AB  divides  both  the  circle  and  its  circumference  into 
equal  parts. 

7.  Definitions.  A  segment  equal  to  one-half  the  circle,  as  the 
segment  AMB,  is  called  a  semicircle.  An  arc  equal  to  half 
a  circumference,  as  the  arc  AMB,  is  called  a  semi-circum- 
ference. 


58 


ELEMENTS  OF  GEOMETRY. 


PROPOSITION  III.— THEOREM. 


8.  In  equal  circles^  or  in  the  same  circle^  equal  angles  at  the 
centre  intercept  equal  arcs  on  the  circumference. 

Let  0,  0\  be  the  centres  of  equal  circles,  and  AOB,  A'O'B', 
equal  angles  at  these  centres ;  then 
the  intercepted  arcs  AB^  A'B',  are 
equal. 

For  the  angle  0'  may  be  super- 
posed upon,  and  made  to  coincide 
with,  its  equal  0.     The  extrem- 
ities of  the  arc  A'B'  will  then  fall  on  the  extremities  of  the 
arc  ABj  and  the  arcs  must  coincide  throughout  and  be  equal, 
since  the  radii  are  equal,     (y.  Proof  of  Proposition  I.) 

9.  Corollary.  Conversely,  in  the  same  circle,  or  in  equal 
circles,  equal  arcs  subtend  equal  angles  at  the  centre. 

10.  Definition.  A  fourth  part  of  a  cir- 
cumference is  called  a  quadrant.  It  is 
evident  from  the  preceding  theorem  that 
a  right  angle  at  the  centre  intercepts  a 
quadrant  on  the  circumference. 

Thus,  two  perpendicular  diameters, 
AOC,  BCD,  divide  the  circumference 
into  four  quadrants,  AB,  BC,  CD,  DA. 


PROPOSITION  IV.— THEOREM. 


11.  In  equal  circles,  or  in  the  same  circle,  equal  arcs  are  sub- 
tended by  equal  chords. 

Let  0,  0',  be  the  centres  of  equal  circles,  and  AB,  A!B\ 
equal  arcs ;  then  the  chords  AB,  A!B',  are  equal. 


BOOK   II. 


59 


For,  drawing  the  radii  to  the  extremities  of  the  arcs,  the 
angles  0  and  0'  are  equal  (Propo-         ^        ^ 
sition  III.,  Corollary),  and  conse-      /^\~/^ 
quently  the  triangles  AOB,  A'0'B\     [      V 
are  equal  (Proposition  YI.,  Book 
I.).     Therefore  AB  =  A'B', 

If  the  arcs  are  in  the  same  circle  the  demonstration  is 
similar. 

12.  Corollary.  Conversely,  in  equal  circles,  or  in  the  same 
circle,  equal  chords  subtend  equal  arcs, 

EXERCISES. 


1.  Theorem. — A  diameter  is  greater  than 
any  other  chord. 


2.  Theorem. —  The  shortest  line  that  can  be 
drawn  from  a  point  within  a  circle  to  the  cir- 
cumference is  a  portion  of  the  diameter  drawn 
through  the  point. 

PROPOSITION  v.— THEOREM. 

13.  In  equal  circles,  or  in  the  same  circle,  the  greater  of  twfr 
unequal  arcs  is  subtended  by  the  greater  chord,  the  arcs  being 
each  less  than  a  semi-circumference. 

Let  the  arc  AC  he  greater  than  the  arc  AB ;  then  the  chord 
AC  iQ  greater  than  the  chord  AB. 

Superpose  the  arc  AB  upon  the 
arc  AC,  placing  centre  upon  cen- 
tre and  A  upon  A;  B  must  fall 
between  A  and  C,  since  AB  is  less 


60 


ELEMENTS   OF  GEOMETRY. 


than  AC.    Draw  now  the  radii  OA,  0J5,  0(7.    In  the  triangles 

AOG  and  AOB  the  angle  AOG  is 

obviously  greater  than  the  angle 

AOB;   therefore,  by  Proposition 

XIV.,  Book  I.,  the  chord  A  a  is 

greater  than  the  chord  AB. 

14.  Corollary.    Conversely,    in 

equal  circles,  or  in  the  same  circle,  the  greater  of  two  unequal 
chords  subtends  the  greater  arc. 

Suggestion,  v.  Proposition  XY.,  Book  I. 

PROPOSITION  VI.— THEOREM. 

15.  The  diameter  perpendicular  to  a  chord  bisects  the  chord 
and  the  arcs  subtended  by  it. 

The  triangles  ACQ,  BCO,  are  equal,  by  Proposition  X., 
Book  I.    Therefore  AC=CB. 

The  triangles  AOD,  BOB,  are  equal,  by 
Proposition  VI.,  Book  I.  Therefore  AD  = 
BD,  and  hence,  by  Proposition  IV.,  Corol- 
lary, the  arc  AD  is  equal  to  the  arc  BD. 

In  the  same  way  we  can  prove  the  arc 
AD'  equal  to  the  arc  BD'.  ^ 

16.  Corollary    I.     The    perpendicular 

erected  at  the  middle  point  of  a  chord  passes  through  the  centre 
of  the  circle,    (v.  Proposition  XVIII.,  Book  I.) 

17.  Corollary  II.  When  two  circumferences  intersect,  the 
straight  line  joining  their  centres  bi- 
sects   their    common    chord  at  right 
angles. 

Suggestion.  Erect  a  perpendicular 
at  the  middle  point  of  the  common 
chord,     (v.  Corollary  I.) 


1 

/    / 

1 
1    / 

^*''J 

BOOK   II.  61 


EXERCISE. 


Theorem. — The  locus  of  the  middle  points  of  a  set  of  parallel 
chords  is  the  diameter  perpendicular  to  the  chords. 


PROPOSITION  VII.— THEOREM. 

18.  In  the  same  circle,  or  in  equal  circles,  equal  chords  are 
equally  distant  from  the  centre  ;  and  of  two  unequal  chords,  the 
less  is  at  the  greater  distance  from  the  centre. 

1st.  Let  AB,  CD,  be  equal  chords;  OE,  OF,  the  perpen- 
diculars which  measure  their  distances 
from  the  centre  0  ;  then  OE  =  OF.  ^ ^^ 

For,  since  the  perpendiculars  bisect         f         ^L^-'^^'^x 
the  chords,  ^^  =  (7i^;  hence  (Propo-      ^LrCT^T^.  I 

sition  X.,  Book  I.)  the  right  triangles        I  ...--"'''/V  / 

AOE  and  COF  are  equal,  and  OE  z=       ^V-v^  y^ 

OF.  M>%^ 

2d.  Let  CG,  AB,  be  unequal  chords  ; 
OE,  OH,  their  distances  from  the  centre ;  and  let  CG  be  less 
than  ^5;  then  OlfyOE. 

For,  since  chord  AB  >  chord  CG,  we  have  arc  AB  >  arc 
CG ;  so  that  if  from  C  we  draw  the  chord  CD  =  AB,  its 
subtended  arc  CD,  being  equal  to  the  arc  AB,  will  be  greater 
than  the  arc  CG.  Therefore  the  perpendicular  OS  will 
intersect  the  chord  CD  in  some  point  I.  Drawing  the  per- 
pendicular OF  to  CD,  we  have,  by  the  first  part  of  the 
demonstration,  OF  =  OE.  But  OH  >  01,  and  01  >  OF 
(Proposition  XYIL,  Book  I.) ;  still  more,  then,  is  OS  >  OF, 
or  OS^OE. 

If  the  chords  be  taken  in  two  equal  circles,  the  demonstra- 
tion is  the  same. 


62 


ELEMENTS   OF   GEOMETRY. 


19.  Corollary.  Conversely^  in  the  same  circle^  or  in  equal 
circles^  chords  equally  distant  from  the  centre  are  equal ;  and 
of  two  chords  unequally  distant  from  the  centre,  that  is  the 
greater  whose  distance  from  the  centre  is  the  less. 


EXERCISE. 

Theorem. — The  least  chord  that  can  be 
drawn  in  a  circle  through  a  given  point  is 
the  chord  perpendicular  to  the  diameter 
through  the  point 

Suggestion,  v.  Proposition  XYII.,  Book  I. 


TANGENTS  AND  SECANTS. 

20.  Definitions.  A  tangent  is  an  indefinite   straight  line 
which  has  but  one  point  in  common 

with  the  circumference;  as  ACB. 
The  common  point,  C,  is  called  the 
point  of  contact,  or  the  point  of  tan- 
gency.  The  circumference  is  also 
said  to  be  tangent  to  the  line  AB 
at  the  point  C, 

A  secant  is  a  straight  line  which 
meets  the  circumference  in  two  points ;  as  JEF. 

Two  circumferences  are  tangent  to  each  other  when  they 
are  both  tangent  to  the  same  straight  line 
at  the  same  point. 

21.  Definition.  A  rectilinear  figure  is  said 
to  be  circumscribed  about  a  circle  when  all 
its  sides  are  tangents  to  the  circumference. 

In  the  same  case,  the  circle  is  said  to  be 
inscribed  in  the  figure. 


BOOK  II.  63 

PROPOSITION  VIII.— THEOREM. 

22.  A  straight  line  cannot  intersect  a  circumference  in  more 
than  two  points. 

For,  if  the  line  could  intersect  the  circumference  in  three 
points,  the  radii  drawn  to  these  points  would  meet  the  line 
at  unequal  distances  from  the  perpendicular  let  fall  from  the 
centre  of  the  circle  upon  the  line,  and  would  be  unequal,  by 
Proposition  XXL,  Book  I. 

PROPOSITION  IX.— THEOREM. 

23.  A  straight  line  tangent  to  a  circle  is  perpendicular  to  the 
radius  drawn  to  the  point  of  contact. 

For  any  other  point  of  the  tangent,                       2)^/ 
as  D,  must  lie  outside  of  the  circle,  and               ^y\\      X 
therefore  the  line  OD^  joining  it  with      a^^   \1  \ 

the  centre,  must  be  greater  than  the             I           q  ] 

radius  0(7,  drawn  to  the  point  of  con-             V  / 

tact.  ^ 

OCiSj  then,  the  shortest  line  that  can 
be  drawn  from  0  to  the  tangent  AB,  and  is  therefore  perpen- 
dicular to  AB,  by  Proposition  XYII.,  Book  I. 

24.  Corollary  I.  A  perpendicular  to  a  tangent  line  drawn 
through  the  point  of  contact  must  pass  through  the  centre  of  the 
circle. 

25.  Corollary  II.  If  two  circumferences  are  tangent  to  each 
other,  their  centres  and  their  point  of  contact  lie  in  the  same 
straight  line. 

Suggestion.  Through  their  point  of  contact  draw  a  line 
perpendicular  to  the  tangent  at  that  point,     (v.  Corollary  I.) 


64 


ELEMENTS   OF   GEOMETRY. 


PROPOSITION  X.— THEOREM. 

26.  When  two  tangents  to  the  same  circle  intersect,  the  dis- 
tances from  their  j)oint  of  intersection  to  their  points  of  contact 
are  equal. 


Eor  the  right  triangles  OAP  and 
OBP  (Proposition  IX.)  are  equal, 
by  Proposition  X.,  Book  I. 


EXERCISES. 

1.  Theorem. — In  any  circumscribed  quadrilateral,  the  sum  of 
two  opposite  sides  is  equal  to  the  sum  of  the  other  two  opposite  sides, 

2.  Theorem. — If  two  circumferences  are  tangent,  and  from  any 
point,  P,  of  the  tangent  at  their  point  of  contact,  tangents  are 
drawn  to  the  two  circles,  the  points  of  contact  of  these  tangents 
are  equally  distant  from  P. 

PROPOSITION  XI.— THEOREM. 

27.  Two  parallels  intercept  equal  arcs  on  a  circumference. 

We  may  have  three  cases : 

1st.  When  the  parallels  AB,  CI),  are 
both  secants,  then  the  intercepted  arcs 
A  G  and  BB  are  equal.  For,  let  OM  be 
the  radius  drawn  perpendicular  to  the 
parallels.  By  Proposition  YI.,  the  point 
M  is  at  once  the  middle  of  the  arc  AMB 
and  of  the  arc  CMD,  and  hence  we  have 

AM  =  BM  and  CM  =  DM, 

whence,  by  subtraction, 

AM—  CM=  BM  —  DM, 

AG  =  BD. 


that  is, 


E 

M 

F 

G/^ 

<^   \  — V 

\d 

(        ^        \ 

A 

V 

\ 

/ 

BOOK  II.  65 

2d.  When  one  of  the  parallels  is  a  secant,  as  AB,  and  the 
other  is  a  tangent,  as  EF  at  Jf,  then  the  intercepted  arcs 
AM  and  BM  are  equal.  For  the  radius  OM  drawn  to  the 
point  of  contact  is  perpendicular  to  the  tangent  (Proposition 
IX.),  and  consequently  perpendicular  also  to  its  parallel  AB  ; 
therefore,  by  Proposition  YI.,  AM  =  BM. 

3d.  When  both  the  parallels  are  tangents,  as  JSF  at  Mj 
and  GH  at  iV,  then  the  intercepted  arcs  MAN"  and  MBN 
are  equal.  For,  drawing  any  secant  AB  parallel  to  the  tan- 
gents, we  have,  by  the  second  case, 

AM  =  BM  and  AN  =  BM, 

whence,  by  addition, 

AM+A]Sr=BM+BW, 
that  is, 

MAN=MBJSr; 

and  each  of  the  intercepted  arcs  in  this  case  is  a  semi-circum- 
ference. 

MEASURE  OF  ANGLES. 

As  the  measurement  of  magnitude  is  one  of  the  principal 
objects  of  geometry,  it  will  be  proper  to  premise  here  some 
principles  in  regard  to  the  measurement  of  quantity  in 
general. 

28.  Definition.  To  measure  sl  quantity  of  any  kind  is  to  find 
how  many  times  it  contains  another  quantity  of  the  same 
kind,  called  the  unit. 

Thus,  to  measure  a  line  is  to  find  the  number  expressing 
how  many  times  it  contains  another  line,  called  the  unit  of 
length,  or  the  linear  unit. 

The  number  which  expresses  how  many  times  a  quantity 
contains  the  unit  is  called  the  numerical  measure  of  that 
quantity. 


66  ELEMENTS   OF   GEOMETRY. 

29.  Definition.  The  ratio  of  two  quantities  is  the  quotient 
arising  from  dividing  one  by  the  oth-er :  thus,  the  ratio  of  A 

to  B  is  ~. 

To  find  the  ratio  of  one  quantity  to  another  is,  then,  to 
find  how  many  times  the  first  contains  the  second;  there- 
fore it  is  the  same  thing  as  to  measure  the  first  by  the 
second  taken  as  the  unit  (28).  It  is  implied  in  the  defini- 
tion of  ratio  that  the  quantities  compared  are  of  the  same 
kind. 

Hence,  also,  instead  of  the  definition  (28),  we  may  say  that 
to  measure  a  quantity  is  to  find  its  ratio  to  the  unit. 

The  ratio  of  two  quantities  is  the  same  as  the  ratio  of  their 
numerical  measures.  Thus,  if  P  denotes  the  unit,  and  if  P 
is  contained  m  times  in  A  and  n  times  in  B^  then 

A mP m 

B~~nP~  n 

30.  Definition.  Two  quantities  are  commensurable  when  there 
is  some  third  quantity  of  the  same  kind  which  is  contained 
a  whole  number  of  times  in  each.  This  third  quantity  is 
called  the  common  measure  of  the  proposed  quantities. 

Thus,  the  two  lines  A  and  B  are  commensurable  if  there 
is  some  line,  (7,  which  is  contained  a 

whole  number  of  times  in  each,  as,      ^'     '     ^     '     '     '     '     ' 
for  example,  7  times  in  J.,  and  4  times      £< — • — « — « — « 
in  B.  0' — « 

The  ratio  of  two   commensurable 
quantities  can,  therefore,  be  exactly  expressed  by  a  number, 

whole  or  fractional  (as  in  the  preceding  example  by  -),  and 

is  called  a  commensurable  ratio. 


BOOK  II.  67 

31.  Definition.  Two  quantities  are  incommensurable  when 
they  have  no  common  measure.  The  ratio  of  two  such 
quantities  is  called  an  incommensurable  ratio. 

If  A  and  B  are  two  incommensurable  quantities,  their  ratio 

is  still  expressed  by  - . 

32.  Problem.  To  find  the  greatest  common  measure  of  two 
quantities.  The  well-known  arithmetical  process  may  be  ex- 
tended to  quantities  of  all  kinds.  Thus,  suppose  AB  and  CD 
are  two  straight  lines  whose  common  measure  is  required. 
Their  greatest  common  measure  can- 
not be  greater  than  the  less  line  CD.      ^' ' ' J"^ 

Therefore  let  CD  be  applied  to  AB  as      Cr—T-pD 

many  times  as  possible,  suppose  three 

times,  with  a  remainder  EB  less  than  CD.  Any  common 
measure  of  AB  and  CD  must  also  be  a  common  measure  of 
CD  and  EB ;  for  it  will  be  contained  a  whole  number  of 
times  in  CD,  and  in  AE^  which  is  a  multiple  of  OD,  and 
therefore  to  measure  AB  it  must  also  measure  the  part 
EB.  Hence  the  greatest  common  measure  of  AB  and  CD 
must  also  be  the  greatest  common  measure  of  CD  and  EB, 
This  greatest  common  measure  of  CD  and  EB  cannot  be 
greater  than  the  less  line  EB ;  therefore  let  EB  be  applied 
as  many  times  as  possible  to  CD,  suppose  twice,  with  a 
remainder  FD.  Then,  by  the  same  reasoning,  the  greatest 
common  measure  of  CD  and  EB^  and  consequently  also 
that  of  AB  and  CD,  is  the  greatest  common  measure  of 
EB  and  FD.  Therefore  let  FD  be  applied  to  EB  as  many 
times  as  possible:  suppose  it  is  contained  exactly  twice  in 
EB  without  remainder;  the  process  is  then  completed,  and 
we  have  found  FD  as  the  required  greatest  common  meas- 
ure. 


68  ELEMENTS   OF   GEOMETRY. 

The  measure  of  each  lin-e,  referred  to  FD  as  the  unit,  will 
then  be  as  follows :  we  have 

EB  =  2FD, 

CD  =  2EB  +FD  =  4FD  +  FD  =  6FD, 

AB  =  SCD  +  EB=  WFD  +  2FD  =  17FD. 

The  proposed  lines  are  therefore  numerically  expressed,  in 

terms  of  the  unit  FD,  by  the  numbers  17  and  5  j  and  their 

..    .    17 
ratio  IS  -— . 
5 

33.  When  the  preceding  process  is  applied  to  two  quanti- 
ties and  no  remainder  can  be  found  which  is  exactly  con- 
tained in  a  preceding  remainder,  however  far  the  process  bo 
contmued,  the  two  quantities  have  no  common  measure ;  that 
is,  they  are  incommensurable,  and  their  ratio  cannot  be  exactly 
expressed  by  any  number,  whole  or  fractional. 

34.  As  the  student  often  has  difficulty  in  realizing  the 
possibility  of  an  incommensurable  ratio,  and  imagines  that  if 
two  lines  are  given  it  must  be  possible  to  take  a  divisor  so 
small  that  it  will  go  exactly  into  each  of  them,  it  seems 
worth  while  to  consider  at  some  length  an  important  exam- 
ple,— namely,  the  ratio  of  the  diagonal  of  a  square  to  one  of 
the  sides. 

Let  the  method  of  (32)  be  applied 
to  finding  the  common  measure  of  the 
diagonal  and  a  side  of  the  square 
ABCD. 

AO  18  clearly  less  than  twice  AB ; 
i.e.,  than  AB  -^  BC.     Lay  off  on  J.  (7 
AB\  equal  to  AB.     Our  problem  is  now 
reduced  to  finding  a  common  measure  of  B'C  and  AB,  or  its 
equal,  CB. 


BOOK  II.  69 

Erect  at  B'  a  perpendicular  B'A'  to  AC.  A'B\  B'C,  and 
A'B  are  all  equal  (y.  Exercise  23,  Book  I.).  If,  then,  we 
lay  off  CB"  equal  to  CB\  B'G  goes  into  BC  twice,  with  a 
remainder  B"A'^  by  which  we  must  proceed  to  divide  B'C. 
But  A' B'C  is  half  a  square,  precisely  similar  to  ABC^  and  in 
performing  the  division  of  B'C,  or  its  equal,  A'B'^  by  A'B"^ 
we  are  merely  repeating,  on  a  smaller  scale,  the  process  just 
performed  in  dividing  BC  loy  B'C  This  will  lead  us  to 
another  repetition,  on  a  still  smaller  scale,  and  so  on  indefi- 
nitely, and  we  shall  never  reach  an  exact  division.  The 
diagonal  and  the  side  of  a  square  have  then  no  common 
divisor,  and  are  absolutely  incommensurable. 

35.  Although  an  incommensurable  ratio  cannot  be  exactly 
expressed  by  a  number,  a  number  can  be  found  by  the  fol- 
lowing method  that  will  approximately  express  it,  and  the 
approximation  may  be  made  as  close  as  we  choose. 

Suppose  that  -  denotes  the  ratio  of  two  incommensurable 

quantities,  A  and  B.  Let  B  be  divided  into  n  equal  parts, 
n  being  some  number  taken  at  pleasure ;  and  then  let  A  be 
divided  by  one  of  these  parts.  Suppose  A  is  found  to  con- 
tain this  divisor  m  times,  with  a  remainder,  which,  of  course, 

is  less  than  the  divisor;  then  -  is  an  approximation  to  the 

value  of  - ,  and  an  approximation  that  may  be  made  as  close 

as  we  please  by  taking  a  sufficiently  great  value  of  n. 

For,  if  X  is  the  magnitude  of  one  of  the  parts  into  which 
B  is  divided,  we  have 

B  =  nXj  while  A^  mx  and  <  (m  -|-  l)x. 
Hence 

^>^and<(^^+^)^: 
B       nx  nx 


70  ELEMENTS  OF  GEOMETEY. 

that  is, 

^  Hes  between  ^  and  ^  +  1 ; 
B  n  n       n 


and  by  increasing  n  we  may  make  -,  which  is  the  difference 

A 

between  two  numbers,  one  less  and  one  greater  than  -,  as 

small  as  we  choose,  and  may  thus  make  the  less  number  - 

n 

A 
as  close  an  approximation  to  the  value  of  -  as  we  please. 

As  a  numerical  example,  take  the  ratio  of  the  diagonal  of 
a  square  to  one  of  the  sides  (34).  If  the  side  is  divided  into 
three  equal  parts,  the  diagonal  will  contain  one  of  these  parts 

four  times,  with  a  remainder  less  than  the  divisor.     -  is  then 

o 

an  approximation,  though  a  very  rough  one,  to  the  value  of 

4         5 

the  ratio  in  question,  which  must  lie  between  -  and  -. 

o  o 

If  the  side  is  divided  into  five  equal  parts,  the  diagonal 

7 
will  contain  seven  of  them,  and  -  is  a  closer  approximation. 

5 

141  1414 

and are  still  closer  approximations. 

100  1000  ^^ 


36.  Definition.  A  proportion  is  an  equation  between  two 
itios.     '] 
equation 


A  A! 

ratios.    Thus,  if  the  ratio  -    is  equal  to  the  ratio  — ,  the 


4=  ^ 
B       B' 

is  a  proportion.     It  may  be  read,  "  Eatio  of  ^  to  jB  equals 
ratio  of  A'  to  5',"  or,  "J.  is  to  B  as  A'  is  to  B'r 


BOOK   II.  71 

A  proportion  is  often  written  as  follows : 

where  the  notation  A  :  B  is  equivalent  to  A  -i-  B.  When 
thus  written,  A  and  B'  are  called  the  extremes^  B  and  A'  the 
means,  and  B'  is  called  a  fourth  proportional  to  A,  B  and  A'  ; 
the  first  terms,  A  and  A\  of  the  ratios  are  called  the  ante- 
cedents ;  the  second  terms,  B  and  B\  the  consequents. 
When  the  means  are  equal,  as  in  the  proportion 
A:B  =  B:C, 

the  middle  term  B  is  called  a  mean  proportional  between  A 
and  Cj  and  C  is  called  a  3f/^^r^  proportional  to  ^  and  j5. 

37.  In  cases  where  it  is  necessary  to  prove  the  equality  of 
incommensurable  ratios,  it  is  usually  best  to  employ  what  is 
called  the  method  of  limits. 
.  38.  Definitions.  A  variable  quantity,  or  simply  a  variable,  is 
a  quantity  whose  value  is  supposed  to  change. 

A  constant  quantity,  or  simply  a  constant,  is  a  quantity 
whose  value  is  fixed. 

The  value  of  a  variable  may  be  changed  at  pleasure,  in 
which  case  it  is  called  an  independent  variable ;  or  it  may  be 
changed  by  changing  at  pleasure  the  value  of  some  other 
variable  or  variables  on  which  it  depends,  and  in  this  case  it 
is  called  a  dependent  variable. 

39.  Definition.  If,  by  changing  in  some  specified  way  the 
variable  on  which  it  depends,  we  can  make  a  dependent  vari- 
able approach  as  near  as  we  please  to  some  given  constant,  but 
can  never  make  the  values  of  the  variable  and  the  constant 
exactly  coincide ;  or,  in  other  words,  if  we  can  make  the  dif- 
ference between  the  variable  and  the  constant  as  small  as  we 
please,  but  cannot  make  it  absolutely  zero,  the  constant  is  called 
the  limit  of  the  variable  under  the  circumstances  specified. 


72  ELEMENTS  OF   GEOMETRY. 

40.  For  example,  consider  the  fraction  -,  where  n  is  sup- 

n 
posed  to  be  an  independent  variable, — Le.^  one  whose  value 

may  be  changed  arbitrarily  and  to  any  extent, — the  fraction 
-  is  then  a  dependent  variable.    By  increasing  n  at  pleasure,  _ 

may  be  made  to  approach  as  near  as  we  please  to  the  value 
zero,  but  can  never  be  made  exactly  equal  to  zero. 

We  say,  therefore,  that  zero  is  the  limit  of  -,  as  n  is  indefi- 

n 

nitely  increased. 

Again,  the  numerical  approximation  to  the  value  of  an 
incommensurable  ratio  (v.  35)  is  a  dependent  variable^  depend- 
ing upon  the  arbitrarily  chosen  number,  n,  of  equal  parts 
into  which  the  denominator  of  the  ratio  is  divided,  and  it 
has  been  shown  to  differ  from  the  actual  value  of  the  ratio 

by  an  amount  less  than  -.     By  increasing  n  at  pleasure  we 

can  make  this  difference  as  small  as  we  please,  but  can  never 
make  it  absolutely  zero,  for  in  that  case  we  should  have 
found  a  common  measure  of  the  incommensurable  numerator 
and  denominator  of  the  given  ratio. 

The  actual  value  of  an  incommensurable  ratio  is,  then,  the 
limit  approached  by  the  approximation  described  in  (35),  as 
n  is  indefinitely  increased. 

41.  The  usefulness  of  the  method  of  limits  flows  entirely 
from  the  following  fundamental  theorem,  the  truth  of  which 
is  almost  axiomatic. 

Theorem. — If  two  variables  dependent  upon  the  same  variable 
are  so  related  that  they  are  always  equals  no  matter  what  value 
is  given  to  the  variable  on  which  they  depend^  and  if  as  the  inde- 
pendent variable  is  changed  in  some  specified  way,  each  of  them 
approaches  a  limit,  the  two  limits  must  be  absolutely  equal. 

For,  in  considering  two  variables  that  are  and  that  always 


BOOK   II.  73 

remain  equal  to  each  other,  we  are  dealing  with  a  single  vary- 
ing value, — i.e.,  their  common  value, — and  it  is  clear  that  a 
single  variable  cannot  be  made  to  approach  as  near  as  we 
please  to  two  diiferent  constant  values  at  the  same  time,  as 
if  it  is  once  brought  between  the  two  values  in  question, 
afterward,  in  approaching  nearer  to  one,  it  must  inevitably 
recede  from  the  other. 

The  student  should  study  this  demonstration  in  connection 
with  that  of  Proposition  XII.,  which  follows. 

PROPOSITION  XII.— THEOREM. 

42.  In  the  same  circle,  or  in  equal  circles,  two  angles  at  the 
centre  are  in  the  same  ratio  as  their  intercepted  arcs. 

Let  AOB  and  AGO  be  two  angles  at  the  centre  of  the  same 

circle,  or  at  the  centres  of 

equal  circles;   AB  and  AC, 

their  intercepted  arcs ;  then 

AOB  ^AB 

AOC      AC'  o 

1st.  Suppose  the  arcs  to  have  a  common  measure,  x,  which 
is  contained  m  times  in  AB  and  n  times  in  A  C     Then  AB  =t 

mx  and  AC  =  nx,  and 

AB rnx m 

AC  nx  n 
Apply  the  measure  x  to  the  arcs  AB  and  AC,  and  draw  radii 
to  the  points  of  division.  The  angle  AOB  is  thus  divided 
into  m  parts,  and  the  angle  AOC  into  n  parts,  all  of  which 
are  equal,  by  Proposition  III.,  Corollary.  Call  any  one  of  these 
smaller  angles  y ;  then  JLO^  =  my  and  AOC  ==  ny,  and 

AOB rny m 

AOC       ny       n' 

Therefore  A^  =  ^, 

AOC       AC' 

or  (v.  36)  AOB  :  AOC  =z  AB  :  AC. 


74  ELEMENTS   OF   GEOMETRY. 

2d.  If  the  arcs  are  incommensurable,  suppose  the  arc  AG 
to  be  divided  into  any  arbitrarily  chosen  number,  w,  of  equal 
parts,  and  let  one  of  the  parts 
be  applied  as  many  times  as 
possible  to  the  arc  AB ;  let  B' 
be  the  last  point  of  division, 
and  draw  the  radius  OB'. 

By  construction,  the  arcs  AB' 
and  AG  are  commensurable.     Therefore,  by  the  proof  above, 

AOB'  ^  A^ 

AOG        AG' 

If,  now,  we  change  n  the  number  of  parts  into  which  A  G 
is  divided,  AB'  and  AOB'  will  change,  and  consequently 
AOB'       ,  AB'     .||     ,  AOB'       .  AB'  .. 

AOG  ""^  AG  ^^^^  '^"^^'-     AOG  ^^^  AG  "''  '^'^  ^""^- 
bles  depending  upon  the  same  variable,  n. 

By  increasing  n  at  pleasure  we  can  make  each  of  the 
equal  parts  into  which  AC  is  divided  as  small  as  we  please, 
and  consequently  the  remainder  B'B,  which  is  necessarily 
less  than  one  of  these  parts,  can  be  made  as  small  as  we 
please.  It  cannot,  however,  be  made  zero,  for  the  arcs  AB 
and  AG  are  incommensurable,  by  hypothesis. 

It  is  clear,  then,  that  if  n  is  indefinitely  increased,  AB'  will 
have  AB  for  its  limitj  and  A  OB'  will  have  A  OB  for  its  limit 
Hence 

-j^^  IS  the  limit  of -j^, 

and 

41  is  the  limit  of  41?', 
AG  AG 

as  n  is  indefinitely  increased. 


BOOK  II.  75 

A  OB'         AB' 
As  the  two  variables  and  — — ,  both  depending  upon 

-4.C/0  Aiy 

n,  are  always  equal,  no  matter  what  the  value  of  w,  and  each 
approaches  a  limit  as  n  is  indefinitely  increased,  the  two 
limits  in  question  are  absolutely  equal  (41).     Hence 

AOB  ^AB 

AOG       AG' 


PROPOSITION  XIII.— THEOREM. 

43.  The  numerical  measure  of  an  angle  at  the  centre  of  a 
circle  is  the  same  as  the  numerical  measure  of  its  intercepted  arc^ 
if  the  adopted  unit  of  angle  is  the  angle  at  the  centre  which 
intercepts  the  adopted  unit  of  arc. 

Let  AOB  be  an  angle  at  the  centre 
0,  and  AB  its  intercepted  arc.  Let 
-10(7  be  the  angle  which  is  adopted  as 
the  unit  of  angle,  and  let  its  intercepted 
arc  AC  he  the  arc  which  is  adopted  as 
the  unit  of  arc.     By  Proposition  XII.,  we  have 

AOB  ^AB 

AOO       AC' 

But  the  first  of  these  ratios  is  the  measure  (28)  of  the  angle 
AOB  referred  to  the  unit  AOC;  and  the  second  ratio  is  the 
measure  of  the  arc  J.J5  referred  to  the  unit  AC.  Therefore, 
with  the  adopted  units,  the  numerical  measure  of  the  angle 
AOB  is  the  same  as  that  of  the  arc  AB. 

44.  Scholium  I.  This  theorem,  being  of  frequent  applica- 
tion, is  usually  more  briefly,  though  less  accurately,  expressed 
t>y  saying  that  an  angle  at  the  centre  is  measured  by  its  inter- 


76  ELEMENTS   OF   GEOMETRY. 

cepted  arc.  In  this  conventional  statement  of  the  theorem, 
the  condition  that  the  adopted  units  of  angle  and  arc  cor- 
respond to  each  other  is  understood ;  and  the  expression  "  is 
measured  by"  is  used  for  "  has  the  same  numerical  measure 
as." 

45.  Scholium  II.  The  right  angle  is,  "by  its  nature,  the  most 
simple  unit  of  angle ;  nevertheless  custom  has  sanctioned  a 
different  unit. 

The  unit  of  angle  generally  adopted  is  an  angle  equal  to 
^  part  of  a  right  angle,  called  a  degree,  and  denoted  by  the 
symbol  °.  The  corresponding  unit  of  arc  is  ^^^  part  of  a 
quadrant  (10),  and  is  also  called  a  degree. 

A  right  angle  and  a  quadrant  are  therefore  both  expressed 
by  90°.  Two  right  angles  and  a  semi-circumference  are  both 
expressed  by  180°.  Four  right  angles  and  a  whole  circum- 
ference are  both  expressed  by  360°. 

The  degree  (either  of  angle  or  arc)  is  subdivided  into 
minutes  and  seconds,  denoted  by  the  symbols '  and  " :  a  minute 
being  -^  part  of  a  degree,  and  a  second  being  -^  part  of  a 
minute.  Fractional  parts  of  a  degree  less  than  one  second 
are  expressed  by  decimal  parts  of  a  second. 

An  angle,  or  an  arc,  of  any  magnitude  is,  then,  numeri- 
cally expressed  by  the  unit  degree  and  its  subdivisions. 
Thus,  for  example,  an  angle  equal  to  ^  of  a  right  angle,  as 
well  as  its  intercepted  arc,  will  be  expressed  by  12°  51' 
25''.714 

46.  Definition.  When  the  sum  of  two  arcs  is  a  quadrant 
(that  is,  90°),  each  is  called  the  complement  of  the  other. 

When  the  sum  of  two  arcs  is  a  semi-circumference  (that 
is,  180°),  each  is  called  the  supplement  of  the  other.  See  (I., 
16). 


BOOK  II.  77 

47.  Definitions.  An  inscribed  angle  is  one  whose  vertex  is 
on  the  circumference  and  whose  sides  are 
chords ;  as  BA  C. 

In  general,  any  rectilinear  figure,  as 
ABCj  is  said  to  be  inscribed  in  a  circle 
when  its  angular  points  are  on  the  circum* 
ference;  and  the  circle  is  then  said  to  be 
circumscribed  about  the  figure. 

An  angle  is  said  to  be  inscribed  in  a  segment  when  its  vertex 
is  in  the  arc  of  the  segment,  and  its  sides  pass  through  the 
extremities  of  the  subtending  chord.  Thus,  the  angle  BAG 
is  inscribed  in  the  segment  BAG. 


PROPOSITION  XIV.— THEOREM. 

48.  An  inscribed  angle  is  measured  by  one-half  its  intercepted 
arc. 

There  may  be  three  cases : 

1st.  Let  one  of  the  sides  AB  of  the  inscribed  angle  BAG 
be   a   diameter;   then  the  measure  of  the 
angle  BAG  is  one-half  the  arc  BG. 

For,  draw  the  radius  OG.  Then,  AOG 
being  an  isosceles  triangle,  the  angles  OAG 
and  OGA  are  equal  (I.,  Proposition  YIII.). 
The  angle  BOG,  an  exterior  angle  of  the 
triangle  AOG,  is  equal  to  the  sum  of  the 
interior  angles  OAG  and  OGA  (I.,  Proposition  XXYI.,  Cor- 
ollary), and  therefore  double  either  of  them.  But  the  an- 
gle BOG,  at  the  centre,  is  measured  by  the  arc  BG  (44) ; 
therefore  the  angle  OAG  is  measured  by  one-half  the  arc 
BG. 

7* 


78 


ELEMENTS   OF   GEOMETRY. 


2d.  Let  the  centre  of  the  circle  fall  within  the  inscribed 
angle  BAG ;  then  the  measure  of  the  angle  BAC  is  one-half 
of  the  arc  BC. 

For,  draw  the  diameter  AD.  The  meas- 
ure of  the  angle  BAB  is,  by  the  first  case, 
one-half  the  arc  BD ;  and  the  measure  of 
the  angle  GAD  is  one-half  the  arc  GD ; 
therefore  the  measure  of  the  sum  of  the 
angles  BAD  and  GAD  is  one-half  the  sum 
of  the  arcs  BD  and  GD ;  that  is,  the  measure  of  the  angle 
BAG  is  one-half  the  arc  BG. 

3d.  Let  the  centre  of  the  circle  fall  without  the  inscribed 
angle  BAG ;  then  the  measure  of  the  angle 
BAG  is  one-half  the  arc  BG. 

For,  draw  the  diameter  AD.  The  meas- 
ure of  the  angle  BAD  is,  by  the  first  case, 
one-half  the  arc  BD ;  and  the  measure  of 
the  angle  GAD  is  one-half  the  arc  GD  ; 
therefore  the  measure  of  the  difference  of 
the  angles  BAD  and  GAD  is  one-half  the 
difference  of  the  arcs  BD  and  GD ;  that  is,  the  measure  of 
the  angle  BAG  is  one-half  the  arc  BG. 

49.  Corollary.  An  angle  inscribed  in  a 
semicircle  is  a  right  angle. 

EXERCISE. 

Theorem. —  The  opposite  angles  of  an  inscribed  quadrilateral 
are  supplements  of  each  other. 


c  D 


BOOK   II. 


79 


PROPOSITION  XV.— THEOREM. 


50.  An  angle  formed  hy  a  tangent  and  a  chord  is  measured 
by  one-half  the  intercepted  arc. 

Let  the  angle  BAC  be  formed  by 
the  tangent  AB  and  the  chord  AC; 
then  it  is  measured  by  one-half  the 
intercepted  arc  AMC. 

For,  draw  the  diameter  AD.  The 
angle  BAD,  being  a  right  angle  (Prop- 
osition IX.),  is  measured  by  one-half 

the  semi-circumference  AMJD ;  and  the  angle  CAD  is  meas- 
ured by  one-half  the  arc  CD;  therefore  the  angle  BAC, 
which  is  the  difference  of  the  angles  BAD  and  CAD,  is 
measured  by  one-half  the  difference  of  AMD  and  CD ;  that 
is,  by  one-half  the  arc  AMC. 

Also,  the  angle  B'AC  is  measured  by  one-half  the  inter- 
cepted arc  ANC.  For,  it  is  the  sum  of  the  right  angle  B'AD 
and  the  angle  CAD,  and  is  measured  by  one-half  the  sum 
of  the  semi-circumference  AND  and  the  arc  CD  ;  that  is,  by 
one-half  the  arc  ANC. 


EXERCISE. 


Prove  Proposition  XY.  by  the  aid  of 
this  figure,  OE  being  a  radius  perpendic- 
ular to  A  C. 

Suggestion.  Complements  of  the  same 
angle  are  equal. 


80 


ELEMENTS   OF   GEOMETRY. 


PROPOSITION  XVI.— THEOREM. 

51.  An  angle  formed  by  two  chords^  intersecting  within  the 
circumference^  is  measured  by  one-half  the  sum  of  the  arcs  inter- 
cepted between  its  sides  and  between  the  sides  of  its  vertical  angle. 

Let  the  angle  AEC  be  formed  by  the  chords  AB,  CD,  in- 
tersecting within  the  circumference ;  then 
will  it  be  measured  by  one-half  the  sum 
of  the  arcs  AG  and  BD^  intercepted  be- 
tween the  sides  of  AEG  and  the  sides  of 
its  vertical  angle  BED, 

For,  join  AD.  The  angle  AEG  is  equal 
to  the  sum  of  the  angles  EDA  and  EAD,  and  these  angles 
are  measured  by  one-half  of  AG  and  one-half  of  BD,  re- 
spectively ;  therefore  the  angle  AEG  is  measured  by  one-half 
the  sum  of  the  arcs  AG  and  BD. 


EXERCISE. 


Prove  Proposition  XYI.  by  the  aid  of  this 
figure,  DF  being  draT\jn  parallel  to  AB.  (v. 
Proposition  XI.) 


PROPOSITION  XVII.— THEOREM. 

52.  An  angle  formed  by  two  secants,  intersecting  without  the 
circumference,  is  measured  by  one-half  the  difference  of  the  inter- 
cepted arcs. 

Let  the  angle  BA  G  be  formed  by  the  se- 
cants AB  and  AG;  then  will  it  be  measured 
by  one-half  the  difference  of  the  arcs  BG 
and  DE. 

For,  join  GD.  The  angle  BDG  is  equal 
to  the  sum  of  the  angles  DAG  and  AGD ; 
therefore  the  angle  A  is  equal  to  the  differ- 


BOOK   II. 


81 


ence  of  the  angles  BDC  and  ACD.  But  these  angles  are 
measured  by, one-half  of  BG  and  one-half  of  JDE  respec- 
tively;  hence. the  angle  A  is  measured  by  one-half  the  differ- 
ence of  BG  aad  DB, 


EXERCISE. 


Prove  Proposition  XYII.  by  the  aid  of  Proposition  XI., 
drawing  a  suitable  figure. 


PROPOSITION  XVIII.-THEOEEM. 

53.  An  angle  formed  by  a  tangent  and  a  secant  is  measured 
by  half  the  difference  of  the  intercepted  arcs. 

For  the  angle  A  is  equal  to  BDG  minus 
ABD,  by  I.,  Proposition  XXYI.,  Corollary. 

54.  Corollary.  An  angle  formed  by  two 
tangents  is  measured  by  half  the  difference  of 
the  intercepted  arcs. 

EXERCISE. 

1.  Prove  Proposition  XYIII.  and  its  Corollary  by  the  aid 
of  Proposition  XI. 

2.  Theorem. — If  through  the  point  of  contact  of  two  tangent 
circles,  two  secants  are  drawn,  the 

chords  joining  the  points  where  the 
secants  cut  the  circles  are  parallel. 
Suggestion.  FED  =  CEa, 

.  • .    DBE  =  GAE. 

Consider,  also,  the  case  where 
the  given  circles  are  internally 
tangent. 


82  ELEMENTS   OF   GEOMETRY. 

PROBLEMS  OF  CONSTRUCTION. 

Heretofore  our  figures  have  been  assumed  to  be  constructed 
under  certain  conditions,  although  methods  of  constructing 
them  have  not  been  given.  Indeed,  the  precise  construction 
of  the  figures  was  not  necessary,  inasmuch  as  they  were  only 
required  as  aids  in  following  the  demonstration  of  principles. 
We  now  proceed,  first,  to  apply  these  principles  in  the  solu- 
tion of  the  simple  problems  necessary  for  the  construction 
of  the  plane  figures  already  treated  of,  and  then  to  apply 
these  simple  problems  in  the  solution  of  more  complex  ones. 

All  the  constructions  of  elementary  geometry  are  effected 
solely  by  the  straight  line  and  the  circumference,  these  being 
the  only  lines  treated  of  in  the  elements ;  and  these  lines  are 
practically  drawrij  or  described,  by  the  aid  of  the  ruler  and 
compasses,  with  the  use  of  which  the  student  is  supposed  to 
be  familiar. 


PROPOSITION  XIX.— PROBLEM. 

65.  To  bisect  a  given  straight  line. 

Let  AB  be  the  given  straight  Kne. 

"With  the  points  A  and  B  as  centres,  and  with  a  radius 
greater  than  the  half  of  AB,  describe  arcs 


D 


intersecting  in  the  two  points  D  and  E. 

Through  these  points  draw  the  straight  line 

DJS,  which  bisects  AB  at  the  point  G.    For, 

J)  and  B  being  equally  distant  from  A  and 

B,  the  straight  line  DE  is  perpendicular  to 

AB    at    its    middle    point   (I.,   Proposition  ^ 

XYIII.). 


-iB 


"jk" 


BOOK  II.  83 

PROPOSITION  XX.— PROBLEM. 

56.  At  a  given  point  in  a  given  straight  line,  to  erect  a  perpen- 
dicular to  that  line. 

Let  AB  be  the  given  line  and  G  the  given  .^  ..^ 

point. 

Take  two  points,  D  and  B,  in  the  line 
and  at  equal  distances  from  C.     With  D     J"d       o       Fb 
and  E  as  centres,  and  a  radius  greater  than 
DCov  GEj  describe  two  arcs  intersecting  in  F.    Then  CF  is 
the  required  perpendicular  (I.,  Proposition  XYIII.). 

57.  Another  solution.  Take  any  point 
O,  without  the  given  line,  as  a  centre. 


0 

from  O  to  C,  describe  a  circumference 


and  with  a  radius  equal  to  the  distance  ^y 


intersecting  J.jB  in  G  and  in  a  second  **• -'' 

point  D.     Draw   the   diameter  DOE^ 

and  join  EG.  Then  EG  will  be  the  required  perpendicular ; 
for  the  angle  EGD^  inscribed  in  a  semicircle,  is  a  right  angle 
(Proposition  XIY.,  Corollary). 

This  construction  is  often  preferable  to  the  preceding,  es- 
pecially when  the  given  point  G  is  at,  or  near,  one  extremity 
of  the  given  line,  and  it  is  not  convenient  to  produce  the  line 
through  that  extremity.  The  point  O  must  evidently  be  so 
chosen  as  not  to  lie  in  the  required  perpendicular. 


,t7l?I?B 


84 


ELEMENTS  OF   GEOMETRY. 


PROPOSITION  XXI.— PROBLEM. 


58.  From  a  given  point  without  a  given  straight  line,  to  lei 

fall  a  perpendicular  to  that  line. 

o 
Let  AB  be  the  given  line  and  G  the 

given  point. 

With  C  as  a  centre,  and  with  a  radius      a — "-^^ — I — r^rr- — b 
sufficiently  great,  describe   an  arc  in- 
tersecting AB  in  D  and  E.     With  D 
and  E  as  centres,  and  a  radius  greater 
than  the  half  of  DE^  describe  two  arcs  intersecting  in  F. 
The  line  CF  is  the  required  perpendicular  (I.,  Proposition 
XYIII.). 

59.  Another  solution.  With  any  point  0  in  the  line  AB  as  a 
centre,  and  with  the  radius  0(7,  describe 

an   arc    GDE   intersecting  AB  in  D. 

With  i)  as  a  centre,  and  a  radius  equal 

to  the  distance  DC,  describe  an  arc  in-  ^ 

tersecting  the  arc  CDE  in  E.    The  line  /% 

CE  is  the  required  perpendicular.    For, 

the  point  I)  is  the  middle  of  the  arc  CJDE^  and  the  radius 

OD  drawn  to  this  point  is  perpendicular  to  the  chord  CE 

(Proposition  YI.). 


li) 


PROPOSITION  XXII.-PROBLEM. 


60.  To  bisect  a  given  arc  or  a  given  angle. 

Ist.  Let  AB  be  a  given  arc. 

Bisect  its  chord  AB  by  a  perpendicular,  as 
in  (55).  This  perpendicular  also  bisects  the 
arc  (Proposition  YI.). 


BOOK   II. 


8) 


2d.  Let  BAC  be  a  given  angle.  With 
^  as  a  centre,  and  with  any  radius,  de- 
scribe an  arc  intersecting  the  sides  of  the 
angle  in  D  and  E,  With  D  and  E  as 
centres,  and  with  equal  radii,  describe 
arcs  intersecting  in  F.  The  straight  line 
AF  bisects  the  arc  DE^  and  consequently  also  the  angle 
BAG, 

61.  Scholium.  By  the  same  construction,  each  of  the  halves 
of  an  arc,  or  an  angle,  may  be  bisected ;  and  thus,  by  succes- 
sive bisections,  an  arc,  or  an  angle,  may  be  divided  into  4,  8, 
16,  32,  etc.,  equal  parts. 


PROPOSITION  XXIII.— PROBLEM. 

62.  At  a  given  point  in  a  given  straight  line,  to  construct  an 
angle  equal  to  a  given  angle. 

Let  J.  be  the  given  point  in  the  straight 
line  AB,  and  0  the  given  angle. 

With  0  as  a  centre,  and  with  any  radius, 
describe  an  arc  MJS^  terminated  by  the  sides 
of  the  angle.  With  A  as  a  centre,  and  with 
the  same  radius,  OM,  describe  an  indefi- 
nite arc  BC.  With  ^  as  a  centre,  and  with 
a  radius  equal  to  the  chord  of  MJ^,  de- 
scribe an  arc  intersecting  the  indefinite  arc  BC  in  D.  Join 
AI).  Then  the  angle  BAD  is  equal  to  the  angle  0.  For 
the  chords  of  the  arcs  MN  and  BD  are  equal;  therefore 
these  arcs  are  equal,  and  consequently  also  the  angles  0 
and  A. 


86  ELEMENTS  OF   GEOMETRY. 

PROPOSITION  XXIV.— PROBLEM. 

63.  Through  a  given  point,  to  draw  a  parallel  to  a  given 
straight  line. 

Let  A  be  the  given  point,  and  JBC  ^ 

the  given  line.  ,\ 

From  any  point  B  in  BO  draw  the  a/. — \ jp 

straight  line  BAD  through  A.    At  the 

point  A,  by  the  preceding   problem,  b      '  ^ 

construct  the  angle  DAE  equal  to  the 

angle  ABC.    Then  AB  is  parallel  to  BC  (I.,  Proposition 

XXIY.,  Corollary  L). 

64,  Scholium.  This  problem  is,  in  practice,  more  accurately 
solved  by  the  aid  of  a  triangle,  con- 
structed of  wood  or  metal.  This 
triangle  has  one  right  angle,  and  its 
acute  angles  are  usually  made  equal 
to  30°  and  60°. 

Let  A  be  the  given  point,  and 
BG  the  given  line.  Place  the  tri- 
angle, BJFD,  with  one  of  its  sides 
in  coincidence  with  the  given  line 

BO.  Then  place  the  straight  edge  of  a  ruler,  MN,  against 
the  side  EF  of  the  triangle.  Now,  keeping  the  ruler  firmly 
fixed,  slide  the  triangle  along  its  edge  until  the  side  ED 
passes  through  the  given  point  A.  Trace  the  line  EAD  along 
the  edge  ED  of  the  triangle ;  then  it  is  evident  that  this  line 
will  be  parallel  to  BG. 

EXERCISE. 

Prohlem.  Two  angles  of  a  triangle  being  given,  to  find  the 
third,  (v.  I.,  Proposition  XXYI.,  and  I.,  Proposition  III.,  Cor- 
ollary I.) 


BOOK   II.  ^7 

PROPOSITION  XXV.— PROBLEM. 

65.  Two  sides  of  a  triangle  and  their  included  angle  being 
given,  to  construct  the  triangle. 

y  & 

Let  6  and  c  be  the  given  sides,  and        Xj^ 

A  their  included  angle. 

Draw  an  indefinite  line  AE,  and 
construct  the  angle  EAF=^A.  On 
AE  take  AC  =  b,  and  on  AF  take 
AB  =  c;  join  BC.  Then  ABC  is 
the  triangle  required ;  for  it  is  formed  with  the  data. 

With  the  data,  two  sides  and  the  included  angle,  only  one 
triangle  can  be  constructed ;  that  is,  all  triangles  constructed 
with  these  data  are  equal,  and  thus  only  repetitions  of  the 
same  triangle  (I.,  Proposition  YL). 

66.  Scholium.  It  is  evident  that  one  triangle  is  always  pos- 
sible, whatever  may  be  the  magnitude  of  the  proposed  sides 
and  their  included  angle. 

PROPOSITION  XXVL— PROBLEM. 

67.  One  side  and  two  angles  of  a  triangle  being  given,  to 
construct  the  triangle. 

Two  angles  of  the  triangle  being  given,      ^^ ^^^^ 

the  third  angle  can  be  found;   and  we  c 

shall  therefore  always  have  given  the 
two  angles  adjacent  to  the  given  side. 
Let,  then,  c  be  the  given  side,  A  and  B 
the  angles  adjacent  to  it. 

Draw  a  line  AB  =  c ;  at  J.  make  an  angle  BAD  =  A,  and 
at  B  an  angle  ABE  =  B.  The  lines  AD  and  BE  intersecting 
in  (7,  we  have  ABC  as  the  required  triangle. 

With  these  data  but  one  triangle  can  be  constructed  (I., 
Proposition  YIL). 


88  ELEMENTS   OF   GEOMETRY. 

68.  Scholium.  If  the  two  given  angles  are  together  equal 
to  or  greater  than  two  right  angles,  the  problem  is  impos- 
sible ;  that  is,  no  triangle  can  be  constructed  with  the  data ; 
for  the  lines  AD  and  BG  will  not  intersect  on  that  side  of 
AB  on  which  the  angles  have  been  constructed. 

PROPOSITION  XXVII.— PROBLEM. 

69.  The  three  sides  of  a  triangle  being  given,  to  construct  the 
triangle, 

a 

Let  a,  bj  and  c  be  the  three  given  sides.        6. . 

Draw  BC  =  a;  with  (7  as  a  centre  and        * 
a  radius  equal  to  b  describe  an  arc ;  with 
.B  as  a  centre  and  a  radius  equal  to  c  de- 
scribe a  second  arc  intersecting  the  first 
in  A.    Then  ABC  is  the  required  triangle. 

With  these  data  but  one  triangle  can  be  constructed  (I., 
Proposition  IX.). 

70.  Scholium.  The  problem  is  impossible  when  one  of  the 
given  sides  is  equal  to  or  greater  than  the  sum  of  the  other 
two  (I.,  Axiom  I.). 

PROPOSITION  XXVIII.— PROBLEM. 

7L  Two  sides  of  a  triangle  and  the  angle  opposite  to  one  of 
them  being  given,  to  construct  the  triangle. 

i a 

1st.  When  the  given  angle  A  is 

acute,  and  the  given  side  a,  oppo-  r/ 

site  to  it  in  the  triangle,  is  less  than  VwiXo 

the  other  sjiven  side  c.  X     7  I  \ 

Construct  an   angle  DAE  =  A.  ^         ^,' '^z,  ^ 

In  one  of  its  sides,  as  AD,  take 

AB  =  c;  with  J5  as  a  centre  and  a  radius  equal  to  a,  describe 


BOOK   II.  89 

an  arc  which  (since  a  <^  c)  will  intersect  AE  in  two  points,  C" 
and  C",  on  the  same  side  of  A.  Join  BC  and  BG",  Then 
either  ABC  or  ABC"  is  the  required  triangle,  since  each  is 
formed  with  the  data ;  and  the  problem  has  two  solutions. 

There  will,  however,  be  but  one  solution,  even  with  these 
data,  when  the  side  a  is  so  much  less  than  the  side  c  as  to  be 
just  equal  to  the  perpendicular  from  B  upon  AE.  For  then 
the  arc  described  from  -S  as  a  centre,  and  with  the  radius  a, 
will  touch  AE  m2i  single  point,  (7,  and  the  required  triangle 
will  be  ABG^  right  angled  at  C. 

2d.  When  the  given  angle  A  is 
either  acute,  right,  or  obtuse,  and 
the  side  a  opposite  to  it  is  greater 
than  the  other  given  side  c. 

The   same   construction   being      "^(""a' ^^^ 

made  as  in  the  first  case,  the  arc 

described  with  j5  as  a  centre,  and  with  a  radius  equal  to  a, 
will  intersect  AE  in  only  one  point,  C,  on  the  same  side  of  A. 
Then  ABC  will  be  the  triangle  required,  and  will  be  the  only- 
possible  triangle  with  the  data. 

The  second  point  of  intersection,  C",  will  fall  in  EA  pro- 
duced, and  the  triangle  ABC  thus  formed  will  not  contain 
the  given  angle. 

72.  Scholium.  The  problem  is  impossible  when  the  given 
angle  A  is  acute  and  the  proposed  side  opposite  to  it  is  less 
than  the  perpendicular  from  B  upon  AE;  for  then  the  arc 
described  from  B  will  not  intersect  AE. 

The  problem  is  also  impossible  when  the  given  angle  is 
right,  or  obtuse,  if  the  given  side  opposite  to  the  angle  is  less 
than  the  other  given  side ;  for  either  the  arc  described  from 
B  would  not  intersect  AEj  or  it  would  intersect  it  only  when 
produced  through  A. 


90  ELEMENTS   OF  GEOMETRY. 

EXERCISE. 

Problem. — The  adjacent  sides  of  a  parallelogram  and  their 
included  angle  being  given,  to  construct  the  parallelogram. 

PROPOSITION  XXIX.— PROBLEM. 

73.  To  find  the  centre  of  a  given  circumference,  or  of  a  given 
arc. 

Take  any  three  points,  A,  B,  and  (7,  in 
the  given  circumference  or  arc,  and  join 
them  by  chords  AB,  BC.  The  perpen- 
diculars erected  at  the  middle  points  of 
these  chords  will  intersect  in  the  required 
centre  (Proposition  YI.,  Corollary  I.). 

74.  Scholium  I.  Only  one  solution  is  possible ;  for,  since  the 
centre  is  equidistant  from  B  and  (7,  it  must  lie  in  the  perpen- 
dicular erected  at  the  middle  point  of  BC  (I.,  Proposition 
XYIII.),  and  since  it  is  equidistant  from  A  and  B,  it  must  lie 
in  the  perpendicular  erected  at  the  middle  point  of  AB  ;  and 
these  perpendiculars  can  have  but  one  point  in  common. 

75.  Scholium  IL  The  same  construction  serves  to  describe 
a  circumference  which  shall  pass  through  three  given  points, 
Aj  B,  C;  or  to  circumscribe  a  circle  about  a  given  triangle, 
ABC;  that  is,  to  describe  a  circumference  in  which  the  given 
triangle  shall  be  inscribed  (47). 

76.  Scholium  III.  It  follows  from  Scholium  1.  that  three 
points  not  in  the  same  straight  line  will  determine  a  circum- 
ference,— i.e,  through  three  points  not  in  the  same  straight 
line  one  circumference,  and  only  one,  can  be  drawn. 

Hence  two  circumferences  cannot  intersect  in  more  than 
two  points;  for  if  they  had  three  points  in  common  they 
would  coincide  throughout. 


BOOK   II. 


91 


PROPOSITION  XXX.— PROBLEM. 

77,  At  a  given  point  in  a  given  circumference^  to  draw  a  tan- 
gent to  the  circumference. 

Let  A  be  the  given  point  in  the  given 
circumference.  Draw  the  radius  OA^  and 
at  A  draw  BAC  perpendicular  to  OA ; 
BG  will  be  the  required  tangent  (Propo- 
sition IX.). 

If  the  centre  of  the  circumference  is 
not  given,  it  may  first  be  found  by  the 
preceding  problem,  or  we  may  proceed 
more  directly  as  follows :  take  two  points, 
D  and  E^  equidistant  from  A  ;  draw  the 
chord  DE^  and  through  A  draw  BAC 
parallel  to  BE.  Since  A  is  the  middle 
point  of  the  arc  BE^  the  radius  drawn 
to  A  will  be  perpendicular  to  BE  (Proposition  YI.),  and  con- 
sequently also  to  BG ;  therefore  jBC  is  a  tangent  at  A, 


PROPOSITION  XXXI.— PROBLEM. 

78.  Through  a  given  point  without  a  given  circle^  to  draw  a 
tangent  to  the  circle. 

Let  0  be  the  centre  of  the  given  circle 
and  P  the  given  point. 

Upon  OP,  as  a  diameter,  describe  a  cir- 
cumference intersecting  the  circumference 
of  the  given  circle  in  two  points,  A  and  A'. 
Draw  PA  and  PA\  both  of  which  will  be 
tangent  to  the  given  circle.  For,  drawing 
the  radii  OA  and  OA',  the  angles  OAP  and  OA'P  are  right 


92 


ELEMENTS   OP   GEOMETRY. 


angles  (Proposition  XIY.,  Corollary) ;  therefore  PA  and  PA^ 
are  tangents  (Proposition  IX.). 

In  practice,  this  problem  is  accurately- 
solved  by  placing  the  straight  edge  of  a 
ruler  through  the  given  point  and  tangent 
to  the  given  circumference,  and  then  tracing 
the  tangent  by  the  straight  edge.  The  pre- 
cise point  of  tangency  is  then  determined 
by  drawing  a  perpendicular  to  the  tangent 
from  the  centre. 

79.  Scholium,  This  problem  always  admits  of  two  solutions. 


EXERCISE. 


Problem. — To  draw  a 
common  tangent  to  two 
given  circles. 

Suggestion.  For  an  ex- 
terior common  tangent,  in 
the  larger  circle  draw  a 
concentric  circle  whose 
radius  is  the  difference  of  the 
radii  of  the  given  circles.  For 
an  interior  common  tangent, 
about  one  of  the  circles  draw 
a  concentric  circle  whose  ra- 
dius is  the  sum  of  the  radii  of 
the  given  circles. 


BOOK   II. 


93 


PROPOSITION  XXXII.— PROBLEM. 

80.   To  inscribe  a  circle  in  a  given  triangle. 

Let  ABC  be  the  given  triangle.  Bisect  any  two  of  its 
angles,  as  B  and  (7,  by  straight  lines  meet- 
ing in  0.  From  the  point  0  let  fall  per- 
pendiculars OX),  OE^  OF,  upon  the  three 
sides  of  the  triangle ;  these  perpendiculars 
will  be  equal  to  each  other  (I.,  Proposi- 
tion XIX.).  Hence  the  circumference  of 
a  circle,  described  with  the  centre  0,  and 
a  radius  =  OD,  will  pass  through  the  three  points  D,  E,  F, 
will  be  tangent  to  the  three  sides  of  the  triangle  at  these 
points  (Proposition  IX.),  and  will  therefore  be  inscribed  in 
the  triangle. 

EXERCISE. 

Problem. —  Upon  a  given  straight  line,  to  describe  a  segment 
which  shall  contain  a  given  angle. 

Suggestion.  Through  one  end  of  the  given 
line  AB  draw  a  line  BC,  making  with  it 
the  given  angle.  The  two  lines  will  be 
one  a  chord  and  the  other  a  tangent. 
Hence  the  centre  of  the  circle  can  be 
found. 


EXERCISES  ON  BOOK  II. 


THEOREMS. 

1.  If  two  circumferences  are  tangent  in- 
ternally, and  the  radius  of  the  larger  is  the 
diameter  of  the  smaller,  then  any  chord  of 
the  larger  drawn  from  the  point  of  contact 
is  bisected  by  the  circumference  of  the 
smaller  {v.  Proposition  XIV.,  Corollary,  and 
Proposition  VI.). 

2.  If  two  equal  chords  intersect  within  a  circle,  the  segments 
of  one  are  respectively  equal  to  the  segments  of  the  other.  What 
is  the  corresponding  theorem  for  the  case  where  the  chords  meet 
when  produced? 

3.  A  circumference  described  on  the  hypotenuse  of  a  right  tri- 
angle as  a  diameter  passes  through  the  vertex  of  the  right  angle. 
(v.  Proposition  XIV.,  Corollary.) 

4.  The  circles  described  on  two  sides  of  a  triangle  as  diameters 
intersect  on  the  third  side. 

Suggestion,  Drop  a  perpendicular  from  the  opposite  vertex  upon 
the  third  side. 

5.  The  perpendiculars  from  the  angles  upon  the  opposite  sides 
of  a  triangle  are  the  bisectors  of  the  angles  of  the  triangle  formed 
by  joining  the  feet  of  the  perpendiculars. 

Suggestion.  On  the  three  sides  of  the  given  triangle  as  diam- 
eters describe  circumferences,  {v.  Exercise  3,  Proposition  XIV., 
and  I.,  Proposition  XXVI.). 

6.  If  a  circle  is  circumscribed  about  an  equilateral  triangle,  the 
perpendicular  from  its  centre  upon  a  side  of  the  triangle  is  equal 
to  one-half  of  the  radius. 

94 


BOOK   II. 


95 


7.  The  portions  of  any  straight  line  which 
are  intercepted  between  the  circumferences 
of  two  concentric  circles  are  equal. 


8.  Two  circles  are  tangent  internally 
at  P,  and  a  chord  AB  of  the  larger  circle 
touches  the  smaller  at  C;  prove  that  PC 
bisects  the  angle  APB. 

Suggestion.  CPQ  =  BCP,  BPQ  = 
BAP,  BCP—BAP  =  APC  (I.,  Propo- 
sition XXVI.,  Corollary). 


9.  If  a  triangle  ABC  is  formed  by  the  intersection  of  three 
tangents  to  a  circumference,  two  of 
which,  A3f  and  AN,  are  fixed,  while 
the  third,  BC,  touches  the  circumfer- 
ence at  a  variable  point  P,  prove  that 
the  perimeter  of  the  triangle  ^^C  is 
constant,  and  equal  to  AM  -\-  AN,  or 
2 AN  (Proposition  X.). 

Also,  prove  that  the  angle  P 0(7  is 
constant. 


10.  If  through  one  of  the  points  of  intersection  of  two  circum- 
ferences a  diameter  of  each  circle  is  drawn,  the  straight  line 
which  joins  the  extremities  of  these  diameters  passes  through  the 
other  point  of  intersection,  and  is  parallel  to  the  line  joining  the 
centres. 

Suggestion.  Draw  the  common  chord  and  the  line  joining  the 
centres,  (v.  Proposition  VI.,  Corollary  II.,  and  Exercise  29, 
Book  I.) 


11.  The  difference  between  the  hy- 
potenuse of  a  right  triangle  and  the 
sum  of  the  other  two  sides  is  equal  to 
the  diameter 'of  the  inscribed  circle. 


I 


96 


ELEMENTS   OF   GEOMETRY. 


12.  A  circle  can  be  entirely  sur- 
rounded by  six  circles  having  the 
same  radius  with  it. 


13.  The  bisectors  of  the  vertical  angles  of  all  triangles  having 
the  same  base  and  equal  vertical  angles  have  a  point  in  common. 

Suggestion,  The  triangles  may  all  be  inscribed  in  the  same 
circle. 

14.  If  the  hypotenuse  of  a  right  triangle  is  double  one  of  the 
sides,  the  acute  angles  of  the  triangle  are  30°  and  60°  respectively. 

15.  If,  from  a  point  whose  distance  from  the  centre  of  a  given 
circle  is  equal  to  a  diameter,  tangents  are  drawn  to  the  circle, 
they  will  make  with  each  other  an  angle  of  60°. 


LOCI. 

16.  Find  the  locus  of  the  centre  of  a  circumference  which  passes 
through  two  given  points,     (v.  I.,  Proposition  XVIII.) 

17.  Find  the  locus  of  the  centre  of  a  circumference  which  is 
tangent  to  two  given  straight  lines,    {v.  I.,  Proposition  XIX.) 

18.  Find  the  locus  of  the  centre  of  a  circumference  which  is 
tangent  to  a  given  straight  line  at  a  given  point  of  that  line,  or 
to  a  given  circumference  at  a  given  point  of  that  circumference. 

19.  Find  the  locus  of  the  centre  of  a  circumference  passing 
through  a  given  point  and  having  a  given  radius. 

20.  Find  the  locus  of  the  centre  of  a  circumference  tangent  to 
a  given  straight  line  and  having  a  given  radius. 

21.  Find  the  locus  of  the  centre  of  a  circumference  of  given 
radius,  tangent  externally  or  internally  to  a  given  circumference. 


BOOK   II. 


97 


22.  A  straight  line  MN,  of  given  length, 
is  placed  with  its  extremities  on  two 
given  perpendicular  lines  ^^,  CD;  find 
the  locus  of  its  middle  point  P  (Exercise 
31,  Book  I.). 


23.  A  straight  line  of  given  length  is  inscribed  in  a  given  circle; 
find  the  locus  of  its  middle  point,     {v.  Proposition  VII.) 


24.  A  straight  line  is  drawn  through  a 
given  point  A^  intersecting  a  given  circum- 
ference in  B  and  C;  find  the  locus  of  the 
middle  point,  P,  of  the  intercepted  chord 
BC. 

Note  the  special  case  in  which  the  point 
A  is  on  the  given  circumference. 


25.  From  any  point  ^  in  a  given  circumference,  a  straight  line 
-4P  of  fixed  length  is  drawn  parallel  to  a  given  line  MN ;  find 
the  locus  of  the  extremity  P,    {v.  I.,  Proposition  XXX.) 


26.  From  one  extremity  -4  of  a  fixed 
diameter  ABy  any  chord  AC  is  drawn, 
and  at  (7  a  tangent  CD.  From  P,  a  per- 
pendicular BD  to  the  tangent  is  drawn, 
meeting  AC  in  P,    Find  the  locus  of  P. 

Suggestion.  (Draw  radius  OC.  v.  I., 
Exercise  28.) 


27.  The  base  BC  of  a  triangle  is  given,  and 
the  medial  line  BE,  from  jB,  is  of  a  given 
length.    Find  the  locus  of  the  vertex  A. 

Suggestion.  Draw  AO  parallel  to  EB.  Since 
BO=BC,  O  is  a  fixed  point;  and  since  AO 
=  2BEy  OA  is  a  constant  distance. 


98  ELEMENTS   OF   GEOMETRY. 


PROBLEMS. 

The  most  useful  general  precept  that  can  be  given,  to  aid  the 
student  in  his  search  for  the  solution  of  a  problem,  is  the  follow- 
ing :  Suppose  the  problem  solved,  and  construct  a  figure  accord- 
ingly ;  study  the  properties  of  this  figure,  drawing  auxiliary  lines 
when  necessary,  and  endeavor  to  discover  the  dependence  of  the 
problem  upon  previously  solved  problems.  This  is  an  analysis 
of  the  problem.  The  reverse  process,  or  synthesis^  then  furnishes 
a  construction  of  the  problem.  In  the  analysis,  the  student's  in- 
genuity will  be  exercised  especially  in  drawing  useful  auxiliary 
lines ;  in  the  synthesis,  he  will  often  find  room  for  invention  in 
combining  in  the  most  simple  form  the  several  steps  suggested  by 
the  analysis. 

The  analysis  frequently  leads  to  the  solution  of  a  problem  by 
the  intersection  of  loci.  The  solution  may  turn  upon  the  deter- 
mination of  the  position  of  a  particular  point.  By  one  condition 
of  the  problem  it  may  appear  that  this  required  point  is  neces- 
sarily one  of  the  points  of  a  certain  line ;  this  line  is  a  locus  of 
the  point  satisfying  that  condition.  A  second  condition  of  the 
problem  may  furnish  a  second  locus  of  the  point ;  and  the  point 
is  then  fully  determined,  being  the  intersection  of  the  two  loci. 

Some  of  the  following  problems  are  accompanied  by  an  analysis 
to  illustrate  the  process. 

28.  To  determine  a  point  whose  distances  from  two  given  inter- 
secting straight  lines,  AB^ 
A^B\  are  given.  •....  c 

Analysis.  The  locus  of 
all  the  points  which  are  at 
a  given  distance  from  AB 
consists  of  two  parallels  to 
ABy  CE,  and  DF,  each  at 
the  given  distance  from 
AB,    The  locus  of  all  the 

points  at  a  given  distance  from  A^B^  consists  of  two  parallels, 
C^E^  and  D^F^^  each  at  the  given  distance  from  A^B\  The  re- 
quired point  must  be  in  both  loci,  and  therefore  in  their  inter- 
section. There  are  in  this  case  four  intersections  of  the  loci,  and 
the  problem  has  four  solutions. 

Construction.  At  any  point  of  AB,  as  A,  erect  a  perpendicular 
CD,  and  make  AC  =  AD  =  the  given  distance  from  AB ;  through 
O  and  2)  draw  parallels  to  AB.  In  the  same  manner,  draw  par- 
allels to  A^B^  at  the  given  distance  A^C  =  A'D\    The  intersec- 


E  0 

\ 

\     / 

A 

\»* 

^^  \ 

/'o 

'\  \ 

y*. 

F 

BOOK   II.  99 

tion  of  the  four  parallels  determines  the  four  points  P^,  Pj?  P^^  ^4» 
each  of  which  satisfies  the  conditions. 


29.  Given  two  perpendiculars,  AB  and  CD, 
intersecting  in  O,  to  construct  a  square,  one 
of  whose  angles  shall  coincide  with  one  of 
the  right  angles  at  O,  and  the  vertex  of  the 
opposite  angle  of  the  square  shall  lie  on  a 
given  straight  line  EF,    (Two  solutions.) 


30.  In  a  given  straight  line,  to  find  a  point  equally  distant  from 
two  given  points  without  the  line. 

31.  To  construct  a  square,  given  its  diagonal. 

32.  Through  a  given  point  P  within  a  given  angle,  to  draw  a 
straight  line,  terminated  by  the  sides  of  the  anglfe,  which  shall 
be  bisected  at  P.     {v.  Exercise  28,  Book  I.) 

33.  Given  two  straight  lines  which  caimot  be  produced  to  their 
intersection,  to  draw  a  third  which  would  pass  through  their 
intersection  and  bisect  their  contained  angle. 

Suggestion.  Find  two  points  equidistant  from  the  two  lines. 
(v.  I.,  Proposition  XIX.) 

34.  Given  the  middle  point  of  a  chord  in  a  given  circle,  to  draw 
the  chord. 

36.  To  draw  a  tangent  to  a  given  circle  which  shall  be  parallel 
to  a  given  straight  line. 

36.  To  draw  a  tangent  to  a  given  circle,  such  that  its  segment 
intercepted  between  the  point  of  contact  and  a  given  straight 
line  shall  have  a  given  length. 

Suggestion.  The  tangent,  the  radius  drawn  to  the  point  of  con- 
tact, and  a  line  drawn  from  the  centre  to  the  end  of  the  tangent 
form  a  right  triangle,  two  of  whose  sides  are  known.  A  simple 
construction  gives  the  hypotenuse. 

In  general  there  are  four  solutions.  Show  when  there  will  be 
but  two ;  also,  when  no  solution  is  possible. 

37.  Through  a  given  point  within  or  without  a  given  circle,  to 
draw  a  straight  line,  intersecting  the  circumference,  so  that  the 
intercepted  chord  shall  have  a  given  length.  (Two  solutions.) 
{v.  Exercise  23  and  Section  78.) 

38.  Construct  an  angle  of  60°,  one  of  120°,  one  of  30°,  one  of  150°, 
one  of  45°,  and  one  of  135°. 


100  ELEMENTS  OF  GEOMETRY. 

39.  Construct  a  triangle,  given  the  base,  the  angle  opposite  to 
the  base,  and  the  altitude. 

Analysis.  Suppose  BAC  to  be  the  re-  ^        s^^ 

quired  triangle.    The  side  BC  being  fixed     ^     7^^. ^\\    ^ 

in  position  and  magnitude,  the  vertex  A  is  / /     ^X;       1 1 

to  be  determined.    One  locus  of  A  is  an  k.^^        '^-vV 

arc  of  a  segment,  described  upon  AB,  con-  ^  ^ 

taining  the  given  angle.    Another  locus 

of  J.  is  a  straight  line  MN  drawn  parallel  to  ^(7,  at  a  distance 
from  it  equal  to  the  given  altitude.  Hence  the  position  of  A  will 
be  found  by  the  intersection  of  these  two  loci,  both  of  which  are 
readily  constructed. 

Limitation.  If  the  given  altitude  were  greater  than  the  perpen- 
dicular distance  from  the  middle  of  ^C  to  the  arc  BAC,  the  arc 
would  not  intersect  the  line  MN,  and  there  would  be  no  solution 
possible. 

The  limits  of  the  data  within  which  the  solution  of  any  prob- 
lem is  possible  should  always  be  determined. 

40.  Construct  a  triangle,  given  the  base,  the  medial  line  to  the 
base,  and  the  angle  opposite  to  the  base. 

41.  With  a  given  radius,  describe  a  circumference,  1st,  tangent 
to  two  given  straight  lines ;  2d,  tangent  to  a  given  straight  line 
and  to  a  given  circumference  ;  3d,  tangent  to  two  given  circum- 
ferences ;  4th,  passing  through  a  given  point  and  tangent  to  a 
given  straight  line  ;  5th,  passing  through  a  given  point  and  tan- 
gent to  a  given  circumference  ;  6th,  having  its  centre  on  a  given 
straight  line,  or  a  given  circumference,  and  tangent  to  a  given 
straight  line,  or  to  a  given  circumference.    (Exercises  19,  20,  21.) 

42.  Describe  a  circumference,  1st,  tangent  to  two  given  straight 
lines,  and  touching  one  of  them  at  a  given  point  (Exercises  17, 
18) ;  2d,  tangent  to  a  given  circumference  at  a  given  point  and 
tangent  to  a  given  straight  line  ;  3d,  tangent  to  a  given  straight 
line  at  a  given  point  and  tangent  to  a  given  circumference  (Exer- 
cise 18) ;  4th,  passing  through  a  given  point  and  tangent  to  a 
given  straight  line  at  a  given  point ;  5th,  passing  through  a  given 
point  and  tangent  to  a  given  circumference  at  a  given  point. 

43.  Draw  a  straight  line  equally  distant  from  three  given  points. 
When  will  there  be  but  three  solutions,  and  when  an  indefinite 

number  of  solutions  ? 

44.  Inscribe  a  straight  line  of  given  length  between  two  given 
circumferences,  and  parallel  to  a  given  straight  line.  {v.  Exer- 
cise 25.) 


BOOK  III. 

PROPORTIONAL   LINES.     SIMILAR   FIGURES. 
THEORY  OF  PROPORTION. 

1.  Definition.  One  quantity  is  said  to  be  proportional  to 
another  when  the  ratio  of  any  two  values,  A  and  JB,  of  the 
first,  is  equal  to  the  ratio  of  the  two  corresponding  values 
A'  and  5',  of  the  second ;  so  that  the  four  values  form  the 
proportion  (II.,  36) 

A:B=A':B% 
or 

B       B'' 

This  definition  presupposes  two  quantities,  each  of  which 
can  have  various  values,  so  related  to  each  other  that  each 
value  of  one  corresponds  to  a  value  of  the  other.  An  exam- 
ple occurs  in  the  case  of  an  angle  at  the  centre  of  a  circle 
and  its  intercepted  arc.  The  angle  may  vary,  and  with  it 
also  the  arc ;  but  to  each  value  of  the  angle  there  corresponds 
a  certain  value  of  the  arc.  It  has  been  proved  (II.,  Proposi- 
tion XII.)  that  the  ratio  of  any  two  values  of  the  angle  is 
equal  to  the  ratio  of  the  two  corresponding  values  of  the 
arc ;  and,  in  accordance  with  the  definition  just  given,  this 
proposition  would  be  briefly  expressed  as  follows :  "  The  angle 
at  the  centre  of  a  circle  is  proportional  to  its  intercepted  arc." 

2.  Definition.  One  quantity  is  said  to  be  reciprocally  propor- 
tional to  another  when  the  ratio  of  two  values,  A  and  B^  of 
the  first,  is  equal  to  the  reciprocal  of  the  ratio  of  the  two 

«*  101 


102  ELEMENTS  OF  GEOMETRY. 

corresponding  values,  A'  and  B\  of  the  second,  so  that  the 
four  values  form  the  proportion 

A\B=B':A\ 
or 

A^B^  ^.  _^  ^ 
B       A'  '   B'' 

For  example,  if  the  product  p  of  two  numbers,  x  and  y,  is 
given,  so  that  we  have 

xy=p,  ^ 

then  X  and  y  may  each  have  an  indefinite  number  of  values, 
but  as  X  increases  y  diminishes.  If,  now,  A  and  B  are  two 
values  of  rr,  while  A'  and  B'  are  the  two  corresponding  values 

of  ?/,  we  must  have 

A  XA'  =  p, 

BX  B'  =  p, 
whence,  by  dividing  one  of  these  equations  by  the  other, 

B^  B'~    ' 
and  therefore 

A  _  J_  ^  ^'  . 
B       A^      A" 
B' 

that  is,  two  numbers  whose  product  is  constant  are  reciprocally 
proportional. 

3.  Let  the  quantities  in  each  of  the  couplets  of  the  pro- 
portion 

j  =  §,ovA:B  =  A':B',  [1] 

be  measured  by  a  unit  of  their  own  kind,  and  thus  expressed 


BOOK    III.  103 

by  numbers  (II.,  28) ;  let  a  and  h  denote  the  numerical  meas- 
ures of  A  and  B,  a'  and  h'  those  of  A'  and  B' ;  then  (II.,  29) 

A^a  A!.  =9l 

B        6'  B'       b" 

and  the  proportion  [1]  may  oe  replaced  by  the  numerical  pro- 
portion 

0  0 

4.  Conversely,  if  the  numerical  measures  a,  6,  a',  h\  of  four 
quantities,  A^  J5,  A\  B',  are  in  proportion,  these  quantities 
themselves  are  in  proportion,  provided  that  A  and  B  are 
quantities  of  the  same  kind,  and  J.'  and  B'  are  quantities  of 
the  same  kind  (though  not  necessarily  of  the  same  kind  as  A 
and  B) ;  that  is,  if  we  have 

a\h  =  a'  :h\ 

we  may,  under  these  conditions,  infer  the  proportion 

A:B  =  A'  :B'. 

5.  Let  us  now  consider  the  numerical  proportion 

a\h  =  a'  :h\ 
Writing  it  in  the  form 


and  multiplying  both  members  of  this  equality  by  hh\  we 
obtain 

ah'  =  a'b, 

whence  the  theorem :  the  product  of  the  extremes  of  a  (numer- 
icaT)  proportion  is  equal  to  the  product  of  the  means. 


4ii»tt. 


104  ELEMENTS   OP   GEOMETRY. 

Corollary.  If  the  means  are  equal,  as  in  the  proportion 
a  :  b  =  b  :  Cj  we  have  b^  =  ac,  whence  b  =  j/ac;  that  is,  a 
mean  proportional  (II.,  36)  between  two  numbers  is  equal  to  the 
square  root  of  their  product. 

6.  Conversely,  if  the  product  of  two  numbers  is  equal  to  the 
product  of  two  others,  either  two  may  be  made  the  extremes,  and 
the  other  two  the  means,  of  a  proportion.    For,  if  we  have  given 

ab'  =  a^bj 
then,  dividing  by  bb',  we  obtain 

^  =  ^'ora:b  =  a':b\ 
b       b 

Corollary.  The  terms  of  a  proportion  may  be  written  in 
any  order  which  will  make  the  product  of  the  extremes 
equal  to  the  product  of  the  means.  Thus,  any  one  of  the 
following  proportions  may  be  inferred  from  the  given  equal- 
ity ab'  =  a'b : 


a 

:b  =a' 

:b\ 

a 

:a'=b 

:6', 

b 

:a=b' 

:a'. 

b 

:b'  =  a 

:a', 

U  '.  a'  =b  '.a,  etc. 
Also,  any  one  of  these  proportions  may  be  inferred  from  any 
other. 

7.  Definitions.  When  we  have  given  the  proportion 

a-.b  =  a'  :b', 
and  infer  the  proportion 

a:  a'  =b  :¥, 
the  second  proportion  is  said  to  be  deduced  by  alternation. 
When  we  infer  the  proportion 

b  :  a  =  b' '.  a', 
this  proportion  is  said  to  be  deduced  by  inversion. 


BOOK   III.  105 

8.  It  is  important  to  observe  that  when  we  speak  of  the 
products  of  the  extremes  and  means  of  a  proportion,  it  is 
implied  that  at  least  two  of  the  terms  are  numbers.  If,  for 
example,  the  terms  of  the  proportion 

A:B=A':  B' 

are  all  lines^  no  meaning  can  be  directly  attached  to  the 
products  ^  X  ^'j  ^  X  A\  since  in  a  product  the  multiplier 
at  least  must  be  a  number. 

But  if  we  have  a  proportion  such  as 

A  \  B  =  m  :  n^ 

in  which  m  and  n  are  numbers,  while  A  and  B  are  any  two 
quantities  of  the  same  kind,  then  we  may  infer  the  equality 
nA  =  mB. 

Nevertheless,  we  shall,  for  the  sake  of  brevity,  often  speak 
of  the  product  of  two  lineSj  meaning  thereby  the  product  of  the 
numbers  which  represent  those  lines  when  they  are  measured  hy  a 
common  unit 

9.  If  A  and  B  are  any  two  quantities  of  the  same  kind, 
and  m  any  number  whole  or  fractional,  we  have,  identically, 

mA A , 

mB~  B' 

that  is,  equimultiples  of  two  quantities  are  in  the  same  ratio  as 
the  quantities  themselves. 

Similarly,  if  we  have  the  proportion 

A:B  =  A':B% 

and  if  m  and  n  are  any  two  numbers,  we  can  infer  the  pro- 
portions 

mA  :  mB  =  nA' :  nB\ 

mA  :  nB  =  mA' :  nB\ 


106  ELEMENTS  OF  GEOMETRY. 

10.  Composition  and  Division.  If  we  have  given  the  pro- 

A       A' 
portion  -  =  — ,  we  have,  by  alternation, 
B       B' 

A'       B'' 
Let  r  be  the  common  value  of  these  two  ratios ;  then 


4,  =  r,and|=r, 


and 


A  =  rA',  and  B  ==  rB\ 

Adding  the  second  equation  to  the  first,  we  have 

A  +  B^KA'+B'), 
or 

A_±B_  ^  A^B_ 

A'  +  B'  A'       B'' 

The  proportions  ^^±|-,  =  A,  and  ^,i|,  =  |  are  said  to 

be  formed  from  the  given  proportion 

A:B^=A':B\hj  composition. 

If  we  subtract  the  equation  B  =  rB'  from  A  =  rA!^  we  have 

A—B  =  r(^Al  —  B'), 
whence,  as  above, 

A  —B  ^A 

A'  —  B'       A" 
and 

A  —B  ^B 
A'  —  B'       B" 

two  proportions  which  are  said  to  be  formed  from  the  given 

proportion 

A  :  B  =  A' : B^hj  division. 

11.  Definition.  A  continued  proportion  is  a  series  of  equal 
ratios,  as 

A:B=zA'  :B'=:A"  :  B"  =  A'"  :  B"'  =  etc. 


i 


BOOK   III.  107 

12.  Let  r  denote  the  common  value  of  the  ratio  in  the 
continued  proportion  of  the  preceding  article ;  that  is,  let 

B       B'       B"       B"' 
then  we  have 

A=::Br,    A'=B'r,    A"  =  B'^r,    A"' =  B'^'r,  etc., 
and,  adding  these  equations, 

A-{.A'  +  A''+  A'"  +  etc.  =  (B-{-B'  +  B"  +  B'"  +  etc.)  r, 
whence 

^  +  ^^+^"+^"^+etc.  ^^^A^A^^^^^, 
BJ^B'^B^'^  B"'  +  etc.  B       B' 

that  is,  the  sum  of  any  number  of  the  antecedents  of  a  continued 
proportion  is  to  the  sum  of  the  corresponding  consequents  as  any 
antecedent  is  to  its  consequent. 

In  this  theorem  the  quantities  J.,  5,  (7,  etc.,  must  all  be 
quantities  of  the  same  kind. 

If,  instead  of  a  continued  proportion,  we  have  an  ordinary 
proportion,  the  theorem  just  proved  obviously  holds  good. 

13.  If  we  have  any  number  of  proportions,  as 

a  '.  h  =^  c  '.  dj 
a'-.h'  =  c' :  d\ 
a"  :  6"  =  c"  :  d'\  etc. ; 
then,  writing  them  in  the  form 

a c     a! c^     a"  c" 

h~  d'    h'~d"     W~dr'       ' 

and  multiplying  these  equations  together,  we  have 

aa'  a" . . . c  d  d' . . . 

hh'h''  ...~  dd'd'\.: 
or 

aaW  ...'.hh'h"  ...  =  cdd'...:dd'd"...] 

that  is,  if  the  corresponding  terms  of  two  or  more  proportions 
are  multiplied  together,  the  products  are  in  proportion. 


108  ELEMENTS  OF  GEOMETRY. 

If  the  corresponding  terms  of  the  several  proportions  are 
equal,  that  is,  ii  a  z=  a'  =  a",  b  =  b' =  V'^  etc.,  then  the  mul- 
tiplication of  two  or  more  proportions  gives 

that  is,  if  four  numbers  are  in  jprojportion^  like  powers  of  these 
numbers  are  in  proportion. 

PEOPOETIONAL  LI:N^ES. 
PROPOSITION  I.— THEOREM. 

14.  A  parallel  to  the  base  of  a  triangle  divides  the  other  two 

sides  proportionally. 

Let  DE  be  a  parallel  to  the  base,  BC,  of  the  triangle  ABC; 

then 

AB:AD=AC  :AE.  ^ 

1st.  Suppose  the  lines  J.J5,  AD,  to  have  #    i^ 

a  common  measure  which  is  contained  m 
times  in  AB  and  n  times  in  AD.     Then       /  [S 

I ^^u' 

AB_m  L A 

AD       n  ^ ^o 

Apply  this  measure  to  AB,  and  through 
the  points  of  division  draw  lines  parallel  to  the  base  BC  o^ 
the  triangle ;  then  through  the  points  of  intersection  of  these 
lines  with  A  G  draw  lines  parallel  to  AB.  The  small  triangles 
thus  formed  are  all  equal,  by  Propositions  XXIX.  and  YII., 
Book  I.  Hence  the  m  parts  into  which  ACiQ  divided  are  all 
equal,  and,  as  AE  contains  n  of  these  parts, 

AC  ^m 
AE       n' 

Therefore 

AB  ^  AC 

AD       AE 


BOOK  III.  109 

2d.  If  AB  and  AD  are  incommensurable,  let  AD  be  divided 
into   any  arbitrarily  chosen   number  n  of 
equal  parts,  and  let  AB  be  divided  by  one  ^ 

of  these  parts.     Let  B'  be  the  last  point  of  A 

division,  5 '5  beinff  of  course  less  than  the  di-  /     \ 

visor.   Through  B'  draw  B' C  parallel  to  DE,  /         \    ~ 

Since  AB'  and  AD  are  commensurable,        /  \ 

A  W         A  C  I  \ 

=  — — -,  and  this  holds  true  no  matter  i//-— -\c" 

AD        AE  t \ 

what  value  may  be  given  to  n.     By  taking 

a  sufficiently  great  value  for  /i,  we  can  make  B'  come  as  near 

''is  we  please  to  B ;  but  we  cannot  make  B'  and  B  coincide, 

since  no  divisor  of  AD  can  divide  AB  without  remainder. 

AB'  AC 

AB'  and  AC,  and   consequently  and  ,  are  then 

^         ^  AD  AE' 

variables  dependent  upon  the  same  variable,  n;  and,  as  we 

have  seen  above,  they  are  equal,  no  matter  what  value  is 

given  to  n.     If  n  is  indefinitely  increased, 


and 


Therefore,  by  the  fundamental  theorem  in  the  Doctrine  of 
Limits  (41,  Book  II.),  these  limits  are  equal,  and  therefore 

AB  ^  AC 

AD~  AE'  \ 

Compare  this  reasoning  with  that  in  II.,  42.  \ 

EXERCISE.  \ 

Show  that  in  Proposition  I.  AD  :  DB  =  AE  :  EC  Q>.  10), 

and  also  that 

AB  ^AD  ^  DB 

AC       AE       EC' 

10 


AB' 
AD 

has  the  limit 

AB 
AD' 

AC 
AE 

has  the  limit 

AC 
AE 

110 


ELEMENTS   OF   GEOMETRY. 


PKOPOSITION  II.— THEOREM. 

15.  Conversely,  if  a  straight  line  divides  two  sides  of  a  tri- 
angle proportionally,  it  is  parallel  to  the  third  side. 

Let  DE  divide  the  sides  AB,  AC,  of  the  tri- 
angle ABC,  proportionally;  then  DE  is  par- 
allel to  BC 

For,  if  DE  is  not  parallel  to  BC,  let  some 
other  line  DE',  drawn  through  D,  be  parallel 
to  BC.     Then,  by  Proposition  I., 

AB:AD=AC:AE'. 

But,  by  hypothesis,  we  have 

AB:AD  =  AC:AE. 


Hence 


AC^ 
AE' 


AG 
AJEf 


whence  it  follows  that  AE'  =  AE,  which  is  impossible  unless 
DE'  coincides  with  DE.    Therefore  DE  is  parallel  to  BC. 


EXERCISE. 

1.  Theorem. — The  line  bisecting  the  vertical  angle  of  a  triangle 
divides  the  base  into  segments  proportional  to  the  adjacent  sides 
of  the  triangle. 

Suggestion.  Through  B  draw 
a  line  parallel  to  the  bisector 
and  extend  the  side  CA  to  meet 
it.  The  triangle  EAB  is  isos- 
celes.    .'.AE  =  AB.     CE  and 

CB  are  divided  proportionally  (Proposition  I.).    Hence  CD  : 
DB  =  CA  :  AB. 

2.  Prove  the  converse  of  Exercise  I.     (v.  Proposition  II.) 


BOOK   III.  Ill 

SIMILAR  POLYGONS. 

16.  Definitions.  Two  polygons  are  similar  when  they  are 
mutually  equiangular  and  have  their  homologous  sides  pro- 
portional. 

In  similar  polygons,  any  points,  angles,  or  lines,  similarly 
situated  in  each,  are  called  homologous. 

The  ratio  of  a  side  of  one  polygon  to  its  homologous  side 
in  the  other  is  called  the  ratio  of  similitude  of  the  polygons. 

PROPOSITION  III.— THEOREM. 

17.  Two  triangles  are  similar  when  they  are  mutually  equi- 
angular. 

Let  ABC,  A'B'C,  be  mutually  ^ 

equiangular  triangles,  in  which  y^i  -4' 

A=A',  B  =  B\  C=  C;   then  /     /  ^f 

these  triangles  are  similar.  /         /  /     / 

For,    superpose    the    triangle     b  c  b!       c 

A'B'C  upon  the  triangle  ABC, 

making  the  angle  A'  coincide  with  its  equal,  the  angle  A,  and 

let  B'  fall  at  h  and  C  at  e.     Since  the  angle  Abe  is  equal  to 

B,  he  is  parallel  to  BC  (I.,  Proposition  XXIY.,  Corollary  I.), 

and  we  have  (Proposition  I.) 

AB:Ah=AC  '.Ac, 
or 

AB\A'B'=^AC'.A'C'. 

If,  now,  we  superpose  A'B'C  upon  ABC,  making  B'  co- 
incide with  B,  we  may  prove,  in  the  same  manner,  that 

AB'.A'B'=BC'.B'C'; 
and,  combining  these  proportions, 

AB   ^  AC_  ^  BC_  PJ-, 

A'B'       A'C       B'C  '-  ^ 

Therefore  the  homologous  sides  are  proportional,  and  the  tri- 
angles are  similar  (16). 


/ 

112  ELEMENTS  OF  GEOMETRY. 

18.  Scholium  I.  The  homologous  sides  lie  opposite  to  equal 
angles. 

19.  Scholium  II.  The  ratio  of  similitude  (16)  of  the  two 
similar  triangles  is  any  one  of  the  equal  ratios  in  the  con- 
tinued proportion  [1]. 

EXERCISE. 

Theorem. — The  altitudes  of  two  similar  triangles  are  to  each 
other  in  the  ratio  of  similitude  of  the  triangles. 


PROPOSITION  IV.— THEOREM. 

20.  Two  triangles  are  similar  when  an  angle  of  the  one  is 
equal  to  an  angle  of  the  other,  and  the  sides  including  these 
angles  are  proportional. 

In  the  triangles  ABC,  A'B'C,                           ^  ^' 

let  A  =  A',  and                                                      yf  /j 

AB         AC  ,                           .X      /  X      / 


A'B'       A'C'  /  Bf       G' 

then  these  triangles  are  similar.         ^  ^ 

For,  place  the  angle  A'  upon  its 

equal  angle  A ;  let  B'  fall  at  h,  and  C  at  c.    Then,  by  the 

hypothesis, 
^  AB^AC 

Ah         Ac 

Therefore  he  is  parallel  to  BC  (Proposition  II.),  and  the  tri- 
angle Ahc  is  similar  to  ABC  (Proposition  III.).  But  Ahc  is 
equal  to  A'B'C;  therefore  A'B'C  is  also  similar  to  ABC. 


BOOK   III.  113 

PROPOSITION  v.— THEOREM. 

21.  Two  triangles  are  similar  when  their  homologous  sides  are 
propoj'tional. 

In  the  triangles  ABC,  A'B'C,  let 

AB_^AG_^BG_,  ry. 

A'B       A'C       B'G"  •-  ■' 

then  these  triangles  are  similar.  ^  ^' 

For,  take  Ah  ==  A!B',  and  Ac  '    /I  /I 

=  ^'(7',  and  join  .6  and  c.  /^    /  /    / 

Abe  is  similar  to  ABC,  by  Prop-         y^'      J         n      c 

osition  lY.     Therefore  b  o 

AB       BO         AB        BC 

-,  or 


Ab         be  '       A'B'        be 

But,  by  hypothesis, 

AB  ^   BC 

A'B'       B'C' 
Hence 

The  triangle  A!B'G'  is  then  equal  to  Abe,  by  I.,  Proposition 
IX.,  and  is  consequently  similar  to  ABC. 

PROPOSITION  VI.— THEOREM. 

22.  If  two  polygons  are  composed  of  the  same  number  of  tri- 
angles similar  each  to  each  and  similarly  placed,  the  polygons 
are  similar. 

Let  the  polygon  ABCD,  /"^^^•^^^^  /t^^^ 

etc.,  be  composed  of  the  tri-     ^/:;.'.. \^    A'k::y.. \n 

angles  J.5  (7,  A  CD,  etc.;  and        \      " '"^^^  V---^'^' 

let  the  polygon  A'B'C'D\  V-^-^""''^ 

etc.,  be   composed   of  the 

triangles  A'B'C\  A'C'D\  etc.,  similar  to  ABC,  ACD,  etc., 

2^  10* 


-^ 


114 


ELEMENTS   OF  GEOMETEY. 


respectively,  and   similarly  placed;  then  the  polygons  are 
similar. 

1st.    The    polygons    are 
mutually  equiangular.  For, 
the  homologous  angles  of 
the    similar    triangles    are 
equal ;  and  any  two  corre- 
sponding angles  of  the  poly- 
gons are  either  homologous  angles  of  two  similar  triangles, 
or  sums  of  homologous  angles  of  two  or  more  similar  trian- 
gles.   Thus,   B  =  B\;   BCD  =  BCA  -f  ACD  =  B'C'A'  + 
A'C'D'=B'C'D' ;  etc. 

2d.  Their  homologous  sides  are  proportional.  For,  from 
the  similar  triangles,  we  have 

AB  ^   BG  ^  AC  ^  CD  ^  AD  ^  DE  ^ 
A'B'       B'C       A'C       CD'       A'D'       D'E'       ^  ^' 
Therefore  the  polygons  fulfil  the  two  conditions  of  similarity 
(16).  \y— 

PEOPOSITION  VII.— THEOREM. 

23.  Conversely,  two  similar  polygons  may  be  decomposed  into 
the  same  number  of  triangles  similar  each  to  each  and  similarly 
placed. 

Let  ABCD,  etc.,  A'B'C'D',  etc.,  be  two  similar  polygons. 
From  two  homologous  ver- 
tices, A  and  A',  let  diag- 
onals be  drawn  in  each 
polygon;  then  the  poly- 
gons will  be  decomposed 
as  required. 

For,  1st.  We  have,  by  the  definition  of  similar  polygons, 

AB         BC 


Angle  B  =  B\  and 


A'B'       B'C 


BOOK   III.  115 

therefore  the  triangles  J.5(7and  A^B'C  are  similar  (Proposi- 
tion lY.). 

2d.  Since  ABC  and  A!B'G'  are  similar,  the  angles  BGA  and 
B'C'A!  are  equal;  subtracting  these  equals  from  the  equals 
BCD  and  B'C'I)\  respectively,  there  remain  the  equals  ACJ) 
and  A' CD',  Also,  from  the  similarity  of  the  triangles  ABC 
and  A'B'C,  and  from  that  of  the  polygons,  we  have 

AC         BC         CD 


A'C       B'C       CD'' 

therefore  the  triangles  ACD  and  A' CD'  are  similar  (Propo- 
sition lY.). 

Thus,  successively,  each  triangle  of  one  polygon  may  be 
shown  to  be  similar  to  the  triangle  similarly  situated  in  the 
other. 

PKOPOSITION  VIII.— THEOREM. 

24.  The  'perimeters  of  two  similar  polygons  are  in  the  same 
ratio  as  any  two  homologous  sides. 

For  we  have  (see  preceding  figures) 


^^         ^^         ^^        etc.; 


A'B'       B'C       CD' 
whence  (12) 

AB  -{-  BC  +    CD  -{-  etc.  ^  AB  ^  BC 
A'B'  +  B'C+  CD'  +  etc.       A'B'       B'C 


etc. 


w 


.1^^ 


116  ELEMENTS  OF  GEOMETRY. 


PROPOSITION  IX.— THEOREM. 

25.  If  a  perpendicular  is  drawn  from  the  vertex  of  the  right 
angle  to  the  hypotenuse  of  a  right  triangle : 

1st.  The  two  triangles  thus  formed  are  similar  to  each  other 
and  to  the  whole  triangle ; 

2d.  The  perpendicular  is  a  mean  proportional  between  the  seg- 
ments of  the  hypotenuse  ; 

3d.  Each  side  about  the  right  angle  is  a  mean  proportional 
between  the  hypotenuse  and  the  adjacent  segment. 

Let  G  be  the  right  angle  of  the  triangle 
ABC,  and  CD  the  perpendicular  to  the 
hypotenuse;  then, 

1st.  The  triangles  ACD  and  CBD  are 
similar  to  each  other  and  to  ABC.  For  the  triangles  ACD 
and  ABC  have  the  angle  A  common,  and  the  right  angles 
ADCj  ACBj  equal;  therefore  they  are  mutually  equiangular, 
and  consequently  similar  (Proposition  III.).  For  a  like 
reason  CBD  is  similar  to  ABCj  and  consequently  also  to 
ACD. 

2d.  The  perpendicular  CD  is  a  mean  proportional  between 

the  segments  AD  and  DB.     For  the  similar  triangles  ACD, 

CBD,  give 

AD:CD  =  CD  :  BD. 

3d.  The  side  J.  (7  is  a  mean  proportional  between  the  hy- 
potenuse AB  and  the  adjacent  segment  AD.  For  the  similar 
triangles,  ACD,  ABC,  give 

AB:AC  =  AC:AD. 

In  the  same  way  the  triangles  CBD  and  ABC  give 

AB:BC  =  BC  :BD. 


BOOK   III.  117 

26.  Scholium.  If  the  lengths  of  the  lines  in  the  figure 
above  are  expressed  in  terms  of  the  same  unit,  the  results 
just  obtained  may  be  written  (5,  Corollary) 

Vlf  =ABX  I>B, 
TO'  =  AB  X  AD, 
'mJ'  =  AB  X  BD. 

27.  Corollary.  If  from  any  point  in  the  circumference  of  a 
circle  a  perpendicular  is  let  fall  upon  a  diameter,  the  perpen- 
dicular is  a  mean  proportional  between  the  segments  of  the  diam- 
eter. 

Suggestion.  Draw  the  chords  A  C  and  CB, 
(v.  11.,  Proposition  XIY.,  Corollary.) 

'<   '  PROPOSITION  X.— THEOREM. 

28.  The  square  of  the  length  of  the  hypotenuse  of  a  right 
triangle  is  the  sum  of  the  squares  of  the  lengths  of  the  other  two 
sides,  the  three  lengths  being  expressed  in  terms  of  the  same  unit. 

Let  ABC  be  right  angled  at  C;  then 

Xg'  =  TO'  +  BU'. 
For,  by  Proposition  IX.,  we  have 

TU'  =  AB  X  AD,  and  277'  =  AB  X  BD, 
the  sum  of  which  is 

■AJ]'  +  'mj'=.AB  X  {AD  ^  BD)  =  AB  X  AB  =  A^. 

29.  Scholium  I.  By  this  theorem,  if  the  numerical  measures 
of  two  sides  of  a  right  triangle  are  given,  that  of  the  third 
is  found.  For  example,  if  ^(7=  3,  BO  =  4;  then  AB  = 
1/ [3^  +  4^  =  5. 

If  the  hypotenuse,  AB,  and  one  side,  AC,  are  given,  we 
have  BC^  =  TB'' —  X0\'  thus,  if  there  are  given  AB  =  b, 
AC  =  3,  then  we  find  BO  =  ^^[5^  —  3']  =  4. 


118 


ELEMENTS   OF   GEOMETRY. 


30.  Scholium  II.  If  J.  0  is  the  diagonal  of  a  square  ABCD, 
we  have,  by  the  preceding  theorem, 


whence 


and,  extracting  the  square  root, 


AC 


-j£  =  \/^=  1.41421  +  ad  inf. 


Since  the  square  root  of  2  is  an  incommensurable  number,  it 
follows  that  the  diagonal  of  a  square  is  incommensurable  with 
its  side.     (v.  II.,  34.) 

31.  Definition.  The  projection  of  a  point 
A  upon  an  indefinite  straight  line  XY  is 
the  foot  P  of  the  perpendicular  let  fall 
from  the  point  upon  the  line. 

The  projection  of  a  finite  straight  line  AB  upon  the  line  XY 
is  the  distance  FQ  between  the  projections  of  the  extremities 
oi  AB. 

If  one  extremity  B  of  the  line  AB  is  in 

A 

the  line-  XY^  the  distance  from  B  to  P  (the 
projection  of  A)  is  the  projection  of  AB  on 
XY;  for  the  point  B  is  in  this  case  its  own 
projectioik 


PM, 


EXERCISES. 


1.  Theorem. — In  any  triangle,  the  square  of  the  side  opposite 
to  an  acute  angle  is  equal  to  the  sum  of  the  squares  of  the  other 
two  sides  diminished  by  twice  the  product  of  one  of  these  sides 
^dnd  the  projection  of  the  other  upon  that  side. 


y^ 


BOOK   III. 


119 


PCy,  Fig.  1,  or 


YiQ.  2. 


Suggestion.  Let  (7  be  the  acute  angle  in  question  in  Fig.  1 
or  Fig  2. 

A       Fig.  1. 

==  XF'  +  (5(7 
-^'^  XF'  +  (PC 

==  2T^  +  TO'  +  :BT'  -  2BC  X  Pc; 
=  Xa'  +  IB^C  —  2BC  X  PC, 

2.  Theorem. — J/i  an  obtuse  angled  triangle, 
the  square  of  the  side  opposite  to  the  obtuse 
angle  is  equal  to  the  sum  of  the  squares  of  the 
other  two  sides,  increased  by  twice  the  product 
of  one  of  these  sides  and  the  projection  of  the 
other  upon  that  side. 

PROPOSITION  XI.— THEOREM. 

32.  If  two  chords  intersect  within  a  circle,  their  segments  are 
reciprocally  proportional. 

For  the  triangles  APB'  and  A'PB  are  mu- 
tually equiangular  (II.,  Proposition  XIY.), 
and    therefore     similar    (Proposition    III.). 

Hence 

AP  :  A'P  =  PB' :  PB. 

33.  Scholium.  If  the  lengths  of  the  lines  in 

question  are  expressed  in  terms  of  the  same  unit,  the  result 
above  can  be  written 

APXPB=  A'P  X  PB', 

and  the  proposition  may  be  stated :  if  through  a  fixed  point 
within  a  circle  any  chord  is  drawn,  the  product  of  the  lengths  of 
its  segments  is  the  same  whatever  its  direction. 


120 


ELEMENTS   OF   GEOMETRY. 


EXERCISE. 

Theorem. — Either  segment  of  the  least  chord 
that  can  he  drawn  through  a  fixed  point  is  a 
mean  proportional  between  the  segments  of  any 
other  chord  drawn  through  that  point,  (v. 
II.,  19,  Exercise.) 


PROPOSITION  XII.— THEOREM. 

34.  If  two  secants  intersect  without  a  circle,  the  whole  secants 
and  their  external  segments  are  reciprocally  proportional. 


For  the  triangles  PAB^  and  PA'B  are  mu- 
tually equiangular  (II.,  Proposition  XIY.), 
and  therefore  similar.     Consequently 

PB  :  PB'  =  PA' :  PA. 


35.  Corollary.  If  a  tangent  and  a  secant 
intersect,  the  tangent  is  a  mean  proportional 
between  the  whole  secant  and  its  external  seg- 
ment. 

Suggestion.  Show  that  the  triangles  PA  T 
and  PTB  are  similar. 

36.  Scholium.  If  the  lengths  of  the  lines 
are  expressed  in  terms  of  the  same  unit, 
the  result  of  (34)  can  be  written  PB  X  PA 

=  PB'  X  PA\  and  Proposition  XII.  can  be  stated :  if  through 
a  fixed  point  without  a  circle  a  secant  is  drawn,  the  product  of 
the  lengths  of  the  whole  secant  and  its  external  segment  has  the 
same  value  in  whatever  direction  the  secant  is  drawn. 


BOOK   III. 


121 


EXERCISE. 

Theorem. — If  from  any  point  on  the  common  chord  of  two 
intersecting  circles^  produced^  tangents  are  drawn  to  the  two 
circles^  the  lengths  of  these  tangents  are  equal. 


PEOBLEMS  OF  CONSTEUCTIOK 
PROPOSITION  XIII.— PROBLEM. 

37.  To  divide  a  given  straight  line  into  any  given  number  of 
equal  parts. 

Let  AB  be  the  given  line.  Through  A  draw 
an  indefinite  line  AX^  upon  which  lay  off  the 
given  number  of  equal  distances,  each  distance 
being  of  any  convenient  length ;  through  M  the 
last  point  of  division  on  AX  draw  MB^  and 
through  the  other  points  of  division  of  AX 
draw  parallels  to  MB^  which  will  divide  AB 
into  the  required  number  of  equal  parts.  This 
follows  from  the  first  part  of  the  proof  of  Proposition  I. 

PROPOSITION  XIV.— PROBLEM. 

38.  To  divide  a  given  straight  line  into  parts  proportional  to 
two  given  straight  lines. 


Let  it  be  required  to  divide  AB 
into  parts  proportional  to  M  and  N, 
From  A  draw  the  indefinite  line  AX^ 
upon  which  lay  off  AG  =  M  and  CD 
=  N.  Join  BB^  and  through  G  draw 
GE  parallel  to  DB.  Then  we  shall 
have  AE  :JEB=AC:GD,  by  Propo- 
sition I. 


I 


122 


ELEMENTS   OF   GEOMETRY. 


EXERCISE. 


Problem. — To  divide  a  given  straight  line 
into  parts  proportional  to  given  straight 
lines. 


...•'5 


M  N 


PROPOSITION  XV.— PROBLEM. 

39.  To  find  a  fourth  proportional  to  three  given  straight  lines. 

Let  it  be  required  to  find  a  fourth  pro- 
portional to  M,  JV,  and  P.  Draw  the  in- 
definite lines  AJT,  AY,  making  an  angle 
with  each  other.  Upon  AJC  lay  off  AB 
=  M,  AD  =  N;  and  upon  ^  Ylay  off  ^(7 
=  P;  join  BG,  and  draw  BE  parallel  to 
BC ;  then  AEi^  the  required  fourth  pro- 
portional. 

For  we  have  (Proposition  I.) 

AB  :  AD=AG:  AE,  or  M :  W=  P  :  AE. 

EXERCISE. 


Prohlem.- 
lines. 


To  find  a  third  proportional  to  tvx)  given  straight 


PROPOSITION  XVI.— PROBLEM. 

40.  To  find  a  mean  proportional  between  two  given  straight 
lines. 

n 

Let  it  be  required  to  find  a  mean  pro- 
portional between  M  and  N.  Upon  an 
indefinite  line  lay  ofl  AB=M,BG  =  JSF; 
upon  AG  describe  a  semi-circumference, 
and  at  B  erect  a  perpendicular,  BD,  to 
AG.  Then  BD  is  the  required  mean  proportional  (Proposi- 
tion IX.,  Corollary). 


BOOK   III. 


123 


41.  Definition.  When  a  given  straight  line  is  divided  into 
two  segments  such  that  one  of  the  segments  is  a  mean  pro- 
portional between  the  given  line  and  the  other  segment,  it  is 
said  to  be  divided  in  extreme  and  mean  ratio. 


Thus,  AB  is  divided  in  extreme  and  mean  ratio     ^ 


H — I 


at  C,ii  AB\AC=:AG  :  CB. 


PROPOSITION  XVII.— PROBLEM. 

42.  To  divide  a  given  straight  line  in  extreme  and  mean  ratio. 

Let  AB  be  the  given  straight  line.  At  B  erect  the  per- 
pendicular BO  equal  to  one-half  of  AB.  With  the  centre  0 
and  radius  OB^  describe  a  circumference, 
and  through  A  and  0  draw  AO  cutting 
the  circumference  first  in  D  and  a  second 
time  in  D'.  Upon  AB  lay  ofl  AG  ==  AD. 
Then  AB  is  divided  at  C  in  extreme  and 
mean  ratio. 

For  we  have  (Proposition  XIL,  Corollary)  ^ 

^J  •  ^^^ 
AD'  :AB=AB:AD  or  AC,  [1] 

whence,  by  division  (10), 

AD'  —  AB:AB  =  AB  —  AC:Aa, 

or,  since  DD'  =20B  =  AB,  and  therefore  AD'  —  AB  =  AD^ 
—  DD'  =  AD=AC, 

AO:AB  =  CB:AC, 
and,  by  inversion  (7), 

AB:AC  =  AC:CB; 
that  is,  AB  is  divided  at  C  in  extreme  and  mean  ratia 


124 


ELEMENTS   OF   GEOMETRY. 


PROPOSITION  XVIII.— PROBLEM. 


43.  On  a  given  straight  line^  to  construct  a  polygon  similar  to 
a  given  polygon. 

Let  it  be  required  to  construct  upon  A'B'  a  polygon  similar 
to  ABCDEF, 

Divide  AB  CDEF  mio  tri- 
angles by  diagonals  drawn 
from  A.  Make  the  angles 
B'A'C  and  A'B'C  equal 
to  BAG  and  ABC  respec- 
tively;   then  the   triangle 

A'B'C  will  be  similar  to  ABC  (Proposition  III.). 
same  manner  construct  the  triangle  A'D'C  similar  to  ADC, 
A' WD'  similar  to  AED,  and  A'E'F'  similar  to  AEF.  Then 
AIB'C'D'E'F'  is  the  required  polygon  (Proposition  YL). 


In  the 


EXERCISES  ON  BOOK  III. 


(1 


THEOREMS. 


1.  If  two  straight  lines  are  intersected  by  any  number  of  par- 
allel lines,  the  corresponding  segments  of  the  two  lines  are  pro- 
portional,    {v.  Proposition  I.) 

2.  The  diagonals  of  a  trapezoid  divide  each  other  into  segments 
which  are  proportional. 

3.  In  a  triangle  any  two  sides  are  reciprocally  proportional  to 
the  perpendiculars  let  fall  upon  them  from  the  opposite  vertices. 

4.  The  perpendiculars  from  two  vertices  of  a  triangle  upon  the 
opposite  sides  divide  each  other  into  segments  which  are  recipro* 
cally  proportional. 

6.  If  the  three  sides  of  a  triangle  are  respectively  perpendicular 
to  the  three  sides  of  a  second  triangle,  the  triangles  are  similar. 

6.  If  ABC  and  A^BO  are  two  triangles  Laving 
a  common  base  BC  and  their  vertices  in  a  line 
AA^  parallel  to  the  base,  and  if  any  parallel  to 
the  base  cuts  the  sides  AB  and  ^  C  in  Z>  and  jE7, 
and  the  sides  A^B  and  A^C  in  D^  and  E^^  then 
DE=^J)'E'  (Proposition  III.). 
""7.  If  two  sides  of  a  triangle  are  divided  propor- 
tionally, the  straight  lines  drawn  from  correspond- 
ing points  of  section  to  the  opposite  angles  intersect 
on  the  line  joining  the  vertex  of  the  third  angle 
and  the  middle  of  the  third  side. 

Suggestion.  Draw  the  line  ADE  through  the  in- 
tersection of  B^C  and  BC^.    B'E'D  and  CED  are 


similar ; 


similar 

BC  ^ 
B'C 

Hence 


EC 
B'E' 

DC 
B'D 


DC 
B'D 
BC 


B'DC^  and  BDC  are 


^   AB^C^  and  ABC  are  similar 


AB 

AB'' 

EC 


B'C 

AB'E'  and  ABE  are  similar ; 
BE 


AB 
AB' 


BE 
B'E' 


B'E'      B'E' 


and  BE  =  EC, 


11*  125 


126 


ELEMENTS   OF  GEOMETRY. 


8.  The  difference  of  the  squares  of  two  sides  of  any  triangle  is 
equal  to  the  difference  of  the  squares  of  the  projections  of  these 
j--^  sides  on  the  third  side  (Proposition  X.). 

^  ^  9.  J[f  from  any  point  in  the  plane  df  a  polj[gon\peitoendiculars 

^"^^  -^alfe^rawn  to  all  the  sides,  the  two  siiml  of\the^qu\res  of  the 
alternate  segments  of  the  sides  are  equkl.  ^ 


^ 


iW 


\J  10.  If  through  a  point  P  in  the  circum- 
ference of  a  circle  two  chords  are  drawn, 
the  chords  and  the  segments  cut  from  them 
by  a  line  parallel  to  the  tangent  at  P  are  re- 
ciprocally proportional. 
Suggestion,  Prove  PAB  and  Pba  similar. 


-^  -41 


11.  If  three  circles  int^fSect,  their  three 
common  chords  pass  through  the  same 
point,    {v.  Proposition  XI.) 


12.  If  two  tangents  are  drawn  to  a  circle 
at  the  extremities  of  a  diameter,  the  por- 
tion of  any  third  tangent  intercepted  be- 
tween them  is  divided  at  its  point  of  con- 
tact into  segments  w^hose  product  is  equal 
to  the  square  of  the  radius. 

Suggestion.  Prove  AOB  a  right  triangle. 


13.  The  perpendicular  from  any  point  of  a  cir- 
cumference upon  a  chord  is  a  mean  proportional 
between  the  perpendiculars  from  the  same  point 
upon  the  tangents  drawn  at  the  extremities  of 
the  chord. 

Suggestion.  PBD  and  PAE  are  similar ;  .  • . 
PB^PD 
PA      PE 


PCE  and  PAD  are  similar 


PA 
PC 


PD 
PE' 


Hence  ^  = 
PA 


PA 
PO' 


BOOK   III. 


127 


14.  If  two  circles  touch  each  other,  secants  drawn  through  their 
point  of  contact  and  terminating  in  the  two  circumferences  are 
divided  proportionally  at  the  point  of  contact,  {v.  II.,  54,  Exer- 
cise 2.)  ^    -        '  ^ 


15.  If  two  circles  are  tangent  exter- 
nally, the  portion  of  their  common 
tangent  included  between  the  points 
of  contact  is  a  mean  proportional  be- 
tween the  diameters  of  the  circles. 

Suggestion.  Show  that  OBO^  is  a  right 
triangle. 

16.  If  a  fixed  circumference  is  cut  by  any  circumference  which 
passes  through  two  fixed  points,  the  com- 
mon chord  passes  through  a  fixed  point. 

Suggestion.  PA  .  PB  =  PC.  PD  =  PT^ 
by  Proposition  XII.  and  Corollary.  Join  P 
with  (7^,  and  show  that  PC^  will  cut  both 
circles  at  the  same  distance  from  P,  and 
will  be  their  common  chord.  "^  ''-::.-.-.-...-=''£  ^ 


^f  :z    X 


LOCI. 


17.  From  a  fixed  point  O,  a  straight  line  OA  is 
drawn  to  any  point  in  a  given  straight  line  JJfJV, 
and  divided  at  P  in  a  given  ratio  m :  n  (ie,  so 
that  OP:PA='m:n);  find  the  locus  of  P.  (v. 
Proposition  II.) 


A' 


18.  From  a  fixed  point  O,  a  straight  line 
OA  is  drawn  to  any  i:)oint  in  a  given  cir- 
cumference, and  divided  at  P  in  a  given 
ratio ;  find  the  locus  of  P. 

Suggestion.  PC  is  a  fixed  length. 


19.  Find  the  locus  of  a  point  whose  distances  from  two  given 
straight  lines  are  in  a  given  ratio.  (The  locus  consists  of  two 
straight  lines.) 


^^r 


f. 


l^i^r-*^  ELEMENTS  OF  GEOMETRY.  / 

yy-  20.  Find  the  locus  of  the  points  which  divide  the  various  chords 
of  a  given  circle  into  segments  whose  product  is  equal  to  a  given 
constant,  k'^  (33,  Exercise). 

21.  Find  the  locus  of  a  point  the  sum  of  whose  distances  from 
two  given  straight  lines  is  equal  to  a  given  constant,  k.  {v,  I., 
Exercise  10.) 

22.  Find  the  locus  of  a  point  the  difference  of  whose  distances 
from  two  given  straight  lines  is  equal  to  a  given  constant,  k. 

Suggestion,  Reduce  it  to  I.,  Proposition  XIX.,  by  drawing  a 
third  line  parallel  to  the  more  distant  of  the  given  lines  at  a 
distance  from  it  equal  to  k, 

^^  PROBLEMS. 

23.  To  divide  a  given  straight  line  into  three  segments.  A,  B, 
and  C,  such  that  A  and  B  shall  be  in  the  ratio  of  two  given 
straight  lines  M  and  iV,  and  B  and  C  shall  be  in  the  ratio  of  two 
other  given  straight  lines  P  and  Q. 

■24.  Through  a  given  point,  to  draw  a  straight  line  so  that  the 
portion  of  it  intercepted  between  two  given  straight  lines  shall 
be  divided  at  the  point  in  a  given  ratio. 

Suggestion.  Through  the  point  draw  a  line  parallel  to  one  of 
the  given  lines,    {v.  II.,  Exercise  32.) 

25.  Through  a  given  point,  to  draw  a  straight  line  so  that  the 
distances  from  two  other  given  points  to  this  line  shall  be  in  a 
given  ratio. 
Suggestion.  Divide  the  line  joining  the  two  other  given  pointe 


26.  To  determine  a  point  whose  distances  from  three  given  in- 
deiinite  straight  lines  shall  be  proportional  to  three  given  straight 
lines.    (Exercise  19.) 

4LJ'  A 


'h^sq 


27.  In  a  given  triangle  ABC,  to  inscribe  a 
square  DEFQ.    (Exercises  6  and  II.,  29.) 


FG 


i:r.~ 


BOOK  m. 


-Hr 


129 


^   28.  Griven  two  circumferences  inter- 

secmng  in  -4,  to  draw  through  A  a 
i  l^ant,  JSACj  such  that  AB  shall  be 

TO  AO  in  Si  given  ratio. 

Suggestion.  Divide  00^  in  the  given 

ratio,    (v.  Exercise  1.) 

29.  To  aescribe  a  circumference  passing  through  two  given 
points  A  and  B   and   tangent  to  a 
given  circumference  O. 

Analysis.  Suppose  ATB  is  the  re- 
guired  circumference  tangent  to  the 
given  circumference  at  T,  and  ACDB 
any  circumference  passing  through  A 
and  B  and  cutting  the  given  circum- 
ference in  C  and  D.  The  common 
chords  AB  and  CD^  and  the  common 
tangent  at  T^  all  pass  through  a  com- 
mon  point   P   (Exercise    16) ;    from 

which  a  simple  construction  may  be  inferred.    There  are  two 
solutions  given  by  the  two  tangents  that  can  be  drawn  from  P. 

<|      30.  To  describe  a  circumference  passing  through  two  given 
^  points  and  tangent  to  a  given  straight  line.    (Two  solutions.) 
{v.  Proposition  XII.,  Corollary.) 

31.  To  describe  a  circumference  passing  through  a  given  point 
and  tangent  to  two  given  straight  lines,    {v.  Exercise  13.) 


\y 


-'lY,'//. 


<^>^xjeJ?i 


>^ir 


BOO 

COMPARISON    AND    MEASUREMENT    OP    THE 
SURFACES    OP    RECTILINEAR   PIGURES. 

1.  Definition.  The  area  of  a  surface  is  its  numerical  meas- 
ure, referred  to  some  other  surface  as  the  unit ;  in  other  words, 
it  is  the  ratio  of  the  surface  to  the  unit  of  surface  (II.,  29). 

The  unit  of  surface  is  called  the  superficial  unit.  The  most 
convenient  superficial  unit  is  the  square  whose  side  is  the 
linear  unit. 

2.  Definition.  Equivalent  figures  are  those  whose  areas  are 

equal. 

PROPOSITION  I.— THEOREM. 

3.  Parallelograms  having  equal  bases  and  equal  altitudes  are 
equivalent. 

Let  ABCD  and  AJECF  be  two 
parallelograms  having  equal  bases 
and  equal  altitudes.  ^  ^ 

Superpose  the  second  upon  the 
first,  making  the  equal  bases  coincide.  Since  the  altitudes 
are  equal,  the  upper  bases  will  lie  in  the  same  straight  line. 
The  triangles  ABE  and  CDF  are  equal  (I.,  Proposition  YL). 
If  the  triangle  CDF  is  taken  from  the  whole  figure,  ABFC, 
the  first  parallelogram  ABCD  is  left;  if  the  equal  triangle 
ABE  is  taken  from  the  same  figure,  the  second  parallelogram 
AECF  is  left.  The  magnitudes  of  the  two  parallelograms 
are  therefore  equal,  and  the  parallelograms  are  equivalent. 

4.  Corollary.  Any  parallelogram  is  equivalent  to  a  rectangle 
having  the  same  base  and  the  same  altitude. 

130 


BOOK   IV. 


131 


PROPOSITION  II.— THEOREM. 
5.  Two  rectangles  having  equal  altitudes  are  to  each  other  as 
their  bases. 

D  O      D  F 


I. 


Let  ABCD  and  AEFD  be  two 
rectangles  having  equal  altitudes ; 
then  are  they  to  each  other  as 
AB  :  AE. 

1.  Suppose  the  bases  have  a  common  measure  which  is 
contained  m  times  in  AB  and  n  times  in  AK     Then  we  have 


AB 
AE 


Apply  this  measure  to  the  two  bases,  and  through  the  points 
of  division  draw  perpendiculars  to  the  bases.  The  two  rec- 
tangles are  thus  divided  into  smaller  rectangles,  all  of  which 
are  equal,  by  I.,  Proposition  XXYIIL,  Corollary,  and  of 
which  ABCD  contains  m  and  AEFD  contains  n.    Then 


ABOD       m 


and  consequently 


AEFD 

ABCD  ^ 
AEFD~ 


n 

AB 
AE' 


c  c 


2.  If  the  bases  are  incommensurable,  divide  AE  in  any 
arbitrarily  chosen  number  n  of 
equal  parts,  and  apply  one  of 
these  parts  to  AB.  Let  B'  be  the 
last  point  of  division,  B'B  being 
of  course  less  than  the  divisor. 

Since  AB'  and  AE  are  commensurable,  we  have  = 

AEFD 

AB' 


£f  B 


AE 


,  and  this  holds,  no  matter  what  the  value  of  n.     If,  now, 


132 


ELEMENTS   OF   GEOMETRY. 


C  O 


n  is  increased  at  pleasure,  we  can  make  J5'J5,  and  consequently 
B'BGG'j  as  small  as  we  please,  but 
cannot  make  them  absolutely  zero. 
Hence,  as  n  is  indefinitely  in- 
creased, AB'  has  AB  for  its  limit, 
AB'C'D  has  ABGD  for  its  Hmit, 


Bf  B 


has    . for  its  limit. 


and 


AEFD 
AB' 


AEFD 


,  „  has  — —  for  its  limit. 
AE         AE 


Therefore,  by  II.,  Theorem^  Doctrine  of  Limits, 


ABCD       AB 


AEFD 


^.     (^v.  IL,  42,  and  III.,  14.) 


6.  Corollary.  Two  rectangles  having  equal  bases  are  to  each 
other  as  their  altitudes. 

Note.  In  these  propositions,  by  "  rectangle"  is  to  be  under- 
stood "  surface  of  the  rectangle." 


v^Vfiv^jfi^^-^ 


V    PBOPOSITION  III.— THEOKEM. 


7.  Any  two  rectangles  are  to  each  other  as  the  products  of  their 
bases  by  their  altitudes. 

Let  B  and  i2'  be  two  rec- 
tangles, k  and  k  their  bases, 
h  and  h'  their  altitudes ;  then 


B 

kXh 
J<fXh'' 

For,  let  >S^  be  a  third  rectan- 
gle, having  the  same  base  k 

BOOK   IV. 


133 


as  the  rectangle  -B,  and  the  same  altitude  h!  as  the  rectangle 
R' ;  then  we  have,  by  Proposition  II.,  Corollary,  and  Propo- 
sition II., 

S       h"        R'       k" 
and  multiplying  these  ratios,  we  find 


R' 


Ji'X  h'' 


8.  Scholium.  It  must  be  remembered  that  by  the  product 
of  two  lines  is  to  be  understood  the  product  of  the  numbers 
which  represent  them  when  they  are  measured  by  the  linear 
unit  (III.,  8). 


PEOPOSITION  IV.— THEOREM. 

9.  The  area  of  a  rectangle  is  equal  to  the  product  of  its  base 
and  altitude. 


Let  R  be  any  rectangle,  k  its  base, 
and  h  its  altitude  numerically  ex- 
pressed in  terms  of  the  linear  unit ; 
and  let  Q  be  the  square  whose  side 
is  the  linear  unit ;  then,  by  Proposition  III., 

R  _k  X  h 


Q      IX  1 


=  k  X  h. 


R 


But  since  Q  is  the  unit  of  surface,  -  =  the  numerical  meas- 
ure,  or  area,  of  the  rectangle,  R,  (1) ;  therefore 


Area  of  R  =  k  X  h. 

12 


134  ELEMENTS  OF  GEOMETRY. 

10.  Scholium  I.  When  the  base  and  altitude  are  exactly 
divisible  by  the  linear  unit,  this  proposition  is  rendered  evi- 
dent by  dividing  the  rectangle  into  squares 

each  equal  to  the  superficial  unit.     Thus,  if 
the  base  contains  7  linear  units  and  the  alti- 
tude 5,  the  rectangle  can  obviously  be  divided 
into  35  squares  each  equal  to  the  superficial 
unit ;  that  is,  its  area  =  5x7.     The  propo- 
sition,  as   above   demonstrated,  is,  however,   more  general, 
and  includes  also  the  cases  in  which  either  the  base  or  the 
altitude,   or  both,   are   incommensurable  with   the  unit  of 
length. 

11.  Scholium  II.  The  area  of  a  square,  being  the  product 
of  two  equal  sides,  is  the  second  power  of  a  side.  Hence  it  is 
that  in  arithmetic  and  algebra  the  expression  "  square  of  a 
number"  has  been  adopted  to  signify  "second  power  of  a 
number." 

We  may  also  here  observe  that  many  writers  employ  the 
expression  "rectangle  of  two  lines"  in  the  sense  of  "product 
of  two  lines,"  because  the  rectangle  constructed  upon  two 
lines  is  measured  by  the  product  of  the  numerical  measures 
of  the  lines. 


PROPOSITION  v.— THEOREM. 

12.  The  area  of  a  parallelogram  is  equal  to  the  product  of  its 
base  and  altitude. 

For,  by  Proposition  I.,  the  parallelogram  is  equivalent  to  a 
rectangle  having  the  same  base  and  the  same  altitude. 


BOOK  IV.         {    I         '^   <    /     '       135 
PROPOSITION  VI.— THEOREM. 

13.  The  area  of  a  triangle  is  equal  to  half  the  product  of  its 
base  and  altitude. 

Let  ABC  be  a  triangle,  k  the  numerical  f- --^ 

measure  of  its  base  BC^h  that  of  its  alti-         •''  ^^^^"^^^ /  •* 
tude  AB^  and  8  its  area  \  then  B^~h o""d 

S=ikXh. 

For,  through  A  draw  AE  parallel  to  CB,  and  through  B  draw 
B£J  parallel  to  CA.  The  triangle  ABC  is  one-half  the  paral- 
lelogram AEBC  (I.,  Proposition  IX.);  but  the  area  of  the 
parallelogram  =  k  y^  h  ;  therefore,  for  the  triangle,  we  have 
S=lk  X  h. 
/   14.  Corollary  I.  A  triangle  is  equivalent  to  one-half  of  any 

i^parallelogram  having  the  same  base  and  the  same  altitude. 

J   15,  Corollary  II.  Triangles  having  equal  bases  and  equal 

\laltitudes  are  equivalent. 

16.  CoROLLA'feY  III.  Triangles  having  equal  altitudes  are  to 
each  other  as  their  bases ;  and  triangles  having  equal  bases  are 
to  each  other  as  their  altitudes. 


X  PROPOSITION  VII.— THEOREM. 

/ 

17.   The  area  of  a  trapezoid  is  equal  to  the  product  of  its  alti^ 

tude  by  half  the  sum  of  its  parallel  bases. 

Let  ABCD  be  a  trapezoid ;  MN=  h,  its 
altitude ;  AD  =  a,  BC  =  b,  its  parallel 
bases ;  and  let  S  denote  its  area ;  then 

S=i(^a  +  b)X  h. 


136 


ELEMENTS   OF   GEOMETRY. 


A    M 


For,  draw  the  diagonal  AC.  The  altitude  of  each  of  the 
triangles  ABC  and  ABC  is  equal  to  h,  and 
their  bases  are  respectively  a  and  b ;  the 
area  of  the  first  is  ^  a  y^  h^  that  of  the 
second  is  ^  6  X  ^ ;  and  the  trapezoid  being 
the  sum  of  the  two  triangles,  we  have  b  n  o 

S=iaXh-]-ibXh  =  i(a  +  b)Xh. 

18.  Scholium.  The  area  of  any  polygon  may  be  found  by 
finding  the  areas  of  the  several  triangles  into  which  it  may 
be  decomposed  by  drawing  diagonals  from  any  vertex. 

The  following  method,  however,  is  usually  preferred,  espe- 
cially in  surveying.  Draw  the  longest 
diagonal  AD  of  the  proposed  polygon 
ABCDEF;  and  upon  AD  let  fall  the 
perpendiculars  BM^  CW,  UP,  FQ. 
The  polygon  is  thus  decomposed  into 
right  triangles  and  right  trapezoids, 
and  by  measuring  the  lengths  of  the 

perpendiculars  and  also  of  the  distances  AM,  MN^  ND,  AQ, 
QP,  PD,  the  bases  and  altitudes  of  these  triangles  and  trape- 
zoids are  known.  Hence  their  areas  can  be  computed  by  the 
preceding  theorems,  and  the  sum  of  these  areas  will  be  the 
area  of  the  polygon. 

i  PROPOSITION  VIII.— THEOREM. 

■'   19.  Similar  triangles  are  to  each  other  as  the  squares  of  their 

homologous  sides. 

Let  ABC,  A'B'C,  be  similar  tri- 
angles; then 

ABC  ^  ZB^ 


BOOK   IV. 
Let  AD  and  A'D'  be  the  altitudes ;  then 


137 


ABC   _    ^AD  X  BC    _  4^        BC 


A'B'C       ^A'D'  X  B'C       A'D'  '^  B'C 


But  the  triangles  ABB  and  A'D'B'  are  similar  (III.,  Propo 
sition  III.)  ]  therefore 

AB        AB 


A'D'       A'B" 


and  from  the  similarity  of  ABC  and  A'B'G\ 


BC        AB 


hence 


and  we  have 


B'C       A'B" 

AB         BC  ^  TB'' 
A'B'  ^  B'G'       A^'' 

ABC  ^  X5' 


EXERCISE. 

\  I 

^    Theorem. — Ti^o  triangles  having  an  angle  of  the  one  equal  to 
an  angle  of  the  other  are  to  each  other  as  the  products  of  the     ^ 
sides  including  the  equal  angles.  L^ 

Suggestion.  Let  ABE  and  ABC  be  the  two 
triangles.     Draw  BE^  and  compare  the  two 
triangles  with  AEB.     (v.    Proposition  YI., 
Corollary  III.) 

12» 


138  ELEMENTS   OF   GEOMETRY. 

PROPOSITION  IX.— THEOREM. 

20.  Similar  polygons  are  to  each  other  as  the  squares  of  their 
homologous  sides. 

Let  ABCDEF,  A' B' C D' E' F' ,  be  two   similar  polygons, 
and  denote  their  surfaces 
by /Sand  >S";  then  b ^  ^      c'       • 

For,  let  the  polygons  be  p 

decomposed  into  homolo- 
gous similar  triangles  (III.,  Proposition  YII.).  The  ratio  of 
the  surfaces  of  any  pair  of  homologous  triangles,  as  ABC  and 
A'B'C,  ACD  and  A' CD',  etc.,  will  be  the  square  of  the  ratio 
of  two  homologous  sides  of  the  polygons  (Proposition  YIII.) ; 
therefore  we  shall  have 

ABO         ACD  ADE  AEF         Tff 


A'B'C       A' CD'       A'D'E'       A'E'F'       Z^' 

Therefore,  by  addition  of  antecedents  and  consequents  (III., 
12). 

ABC  +  ACD   +   ADE   +    AEF   ^S^  Z^ 
A'B'C  +  A' CD'  +  A'D'E'  +  A'E'F'       S'      ATB"^' 


PROPOSITION  X.— THEOREM. 

21.  The  square  described  upon  the  hypotenuse  of  a  right  tri- 
angle is  equivalent  to  the  sum  of  the  squares  described  on  the 
other  two  sides. 


BOOK   IV. 


139 


Let  the  triangle  ABC  be  right  angled  at  C;  then  the 
square  AIT,  described  upon  the 
hypotenuse,  is  equal  in  area  to 
the  sum  of  the  squares  AF  and 
£D,  described  on  the  other  two 
sides. 

For,  from  C  draw  CP  perpen- 
dicular to  AB  and  produce  it  to 
meet  KH  in  Z.  Join  CK,  BG. 
Since  ACF  and  ACB  are  right 
angles,  CF  and  CB  are  in  the 
same  straight  line  (I.,  Proposition 
lY.) ;  and  for  a  similar  reason  AG  and  CD  are  in  the  same 
straight  line. 

In  the  triangles  OAK,  GAB,  we  have  AK  equal  to  AB, 
being  sides  of  the  same  square;  AC  equal  to  AG,  for  the 
same  reason;  and  the  angles  CAK,  GAB,  equal,  being  each 
equal  to  the  sum  of  the  angle  GAB  and  a  right  angle;  there- 
fore  these  triangles  are  equal  (L,  Proposition  VI.). 

The  triangle  CAK  and  the  rectangle  AL  have  the  same 
base  AK;  and  since  the  vertex  C  is  upon  LP  produced,  they 
also  have  the  same  altitude;  therefore  the  triangle  CAK  is 
equivalent  to  one-half  the  rectangle  AL  (Proposition  VI 
Corollary  I.).  ' 

The  triangle  GAB  and  the  square  AF  have  the  same  base 
AG;  and  smce  the  vertex  B  is  upon  FG  produced,  they  also 
have  the  same  altitude ;  therefore  the  triangle  GAB  is  equiva- 
lent to  one-half  the  square  AF  (Proposition  VI.,  Corollary  I ) 

But  the  triangles  CAK,  GAB,  have  been  shown  to  be  equal  • 
therefore  the  rectangle  AL  is  equivalent  to  the  square  AF 

In  the  same  way  it  is  proved  that  the  rectangle  BL  is 
equivalent  to  the  square  BD. 


140 


ELEMENTS   OF   GEOMETRY. 


Therefore  the  square  AH^  which  is  the  sum  of  the  rec- 
tangles AL  and  BL,  is  equivalent 
to  the  sum  of  the  squares  AF  and 
BD. 

22.  Scholium.  This  theorem  is 
ascribed  to  Pythagoras  (born 
about  600  B.C.),  and  is  commonly 
called  the  Pythagorean  Theorem. 
The  preceding  demonstration  of 
it  is  that  which  was  given  by 
Euclid,  in  his  Elements  (about 
300  B.C.). 

It  is  important  to  observe  that  we  may  deduce  the  same 
result  from  the  numerical  relation  XS^  ==:AC^  -\-  ^FU^  already 
established  in  III.,  Proposition  X.  For,  since  the  measure 
of  the  area  of  a  square  is  the  second  power  of  the  number 
which  represents  its  side,  it  follows  directly  from  this  numer- 
ical relation  that  the  area  of  which  AB^  is  the  measure  is 
equal  to  the  sum  of  the  areas  of  which  AC  and  BTf  are  the 
measures. 


EXERCISES. 


1.  Theorem. — If  the  three  sides  of  a  right  triangle  be  taken  as 
the  homologous  sides  of  three  similar  polygons. constructed  upon 
them,  then  the  polygon  constructed  upon  the  hypotenuse  is  equiva- 
lent to  the  sum  of  the  polygons  constructed  upon  the  other  two 
sides,     (v.  Proposition  IX.) 

2.  Theorem. —  The  squares  on  the  sides  of  a  right  triangle  are 
proportional  to  the  segments  into  which  the  hypotenuse  is  divided 
by  a  perpendicular  let  fall  from  the  vertex  of  the  right  angle, 
(v.  Figure,  Proposition  X.) 


BOOK   IV.  141 

PEOBLEMS  OF  C0:N^STEUCTI0K. 
PROPOSITION   XI.— PROBLEM. 

23.  To  construct  a  triangle  equivalent  to  4:  given  polygon. 

Let  ABCDEF  be  the  given  polygon. 
'  Take  any  three  consecutive  vertices, 
as  J.,  5,  (7,  and  draw  the  diagonal  AC' 
Through  B  draw  BP  parallel  to  AO 
Wk  meeting  DC  produced  in  P;  join  AP. 

The   triangles  APG,  ABC,  have   the 
same  base  AG ;  and  since  their  vertices, 

P  and  5,  lie  on  the  same  straight  line  BP  parallel  to  AG^ 
they  also  have  the  same  altitude ;  therefore  they  are  equiva- 
lent. Therefore  the  pentagon  APDEF  is  equivalent  to  the 
hexagon  ABGDEF.  Now,  taking  any  three  consecutive  ver- 
tices of  this  pentagon,  we  shall,  by  a  precisely  similar  con- 
struction, find  a  quadrilateral  of  the  same  area ;  and,  finally, 
by  a  similar  operation  upon  the  quadrilateral  we  shall  find  a 
triangle  of  the  same  area. 

Thus,  whatever  the  number  of  the  sides  of  the  given  poly- 
gon, a  series  of  successive  steps,  each  step  reducing  the  num- 
ber of  sides  by  one,  will  give  a  series  of  polygons  of  equal 
areas,  terminating  in  a  triangle. 

PROPOSITION  XII.— PROBLEM. 

24.  To  construct  a  square  equivalent  to  a  given  parallelogram 
or  to  a  given  triangle.  ^  ^ 

1st.  Let  J.  (7  be  a  given  parallelogram,  k  its      \  ^i  \ 


base,  and  h  its  altitude. 

Find  a  mean  proportional  x  between  h  and  j y 

k,  by  III.,  40.    The  square  constructed  upon 

X  will  be  equivalent  to  the  parallelogram,  since  x^  =  h  y^  k. 


\J 


142 


ELEMENTS   OF   GEOMETRY. 


n 


2d.  Let  ABC  be  a  given  triangle,  a  its  base,  and  h  its 
altitude. 

A 

Find  a  mean  proportional  x  between  a 
and  ih;  the  square  constructed  upon  x 
will  be  equivalent  to  the  triangle,  since     ^  a         u 

=  a  X  i  h  =  i  ah.  I 1 

25.  Scholium.  By  means  of  this  problem 
and  the  preceding,  a  square  can  be  found  equivalent  to  any- 
given  polygon. 


Pi- 


PEOPOSITION  XIII.-PROBLEM. 

26.  To  construct  a  square  equivalent  to  the  sum  of  two  or  more 
given  squares,  or  to  the  difference  of  two  given  squares. 

1st.  Let  m,  w,  ^,  q,  be  the  sides  of  given 
squares. 

Draw  AB  =  m,  and  BC  =  n,  perpendic- 
ular to  each  other  at  B;  join  AC.  Then 
(Proposition  X.)  ZU^  ^  m^  -\-  n\ 

Draw  CD  =  p  perpendicular  to  A  (7,  and 
join  AD.  Then  AJf  =  TU'  +  p' =  m'  + 
n^  +  _p^ 

Draw  DB  =  q  perpendicular  to  AD,  and 
join  AK     Then  TE^  =  Jlf  -\-  q' =  m,' + 
n^  _|_  p^  _|_  ^2     Therefore  the  square   con- 
structed upon  AE  will  be  equivalent  to  the   sum  of  the 
squares  constructed  upon  m,  7i,  p,  q. 

In  this  manner  may  the  areas  of  any  number  of  given 
squares  be  added. 

2d.  Construct  a  right  angle  ABC,  and  lay 
off  BA  =  n.  With  the  centre  A  and  a  radius 
=  m,  describe  an  arc  cutting  BCinC.    Then 


u 


BOOK   IV. 


143 


:ro'  =  zu^  -  zs' 


m' 


therefore   the   square   con- 


M 


structed  upon  BC  will  be  equivalent  to  the  difference  of  the 
squares  constructed  upon  m  and  n. 

EXERCISE. 

Prohlem.^Upon  a  given  straight  line,  to  construct  a  rectangle 
equivalent  to  a  given  rectangle. 

\  PBOPOSITION  XIV.-PROBLEM. 

27.  To  construct  a  rectangle,  having  given  its  area  and  the 
sum  of  two  adjacent  sides. 

Let  MN  be  equal  to  the  given  sum  of 

the  adjacent  sides  of  the  required  rec-  ^ .^ 

tangle ;  and  let  the  given  area  be  that  of      i,,..^.^::C:r\9 
the  square  whose  side  is  AB.  ''  ^ — ^^;--— 

Upon  MJSr  as  a  diameter  describe  a 
semicircle.     At  M  erect  MP  =  AB  per- 
pendicular to  MN,  and  draw  PQ  parallel  to  MN,  intersecting 
he  circumference  in  Q.     From  Q  let  fall  QR  perpendicular 
to  MN;  then  MR  and  RN  are  the  base  and  altitude  of  the 

PROPOSITION  XV.-PEOBLEM. 

28.  To  find  two  straight  lines  in  the  ratio  of  the  areas  of  two 
given  polygons,  • 

Let  squares  be  found  equal  in  area  to  o 

the  given  polygons  respectively  (23  and  /T^v. 

24).     Upon  the  sides  of  the  right  angle      '^ [       ^_j, 
ACB,  take  CA  and  CB  equal  to  the  sides 
of  these  squares,  join  AB,  and  let  fall  CD  perpendicular  to 


B    N 


144  ELEMENTS  OF  GEOMETRY. 

AB.    Then,  by  (III.,  26),  we  have  lU'  =  AD  X  AB,  VF^ 
=  JDB  X  AS-    Hence 

IV  ^  AD . 
VB'       DB' 


therefore  AD  and  DB  are  in  the  ratio  of  the  areas  of  the 
given  polygons. 


EXERCISE. 


Problem. — To  find  a  square  which  shall  he  to  a  given  square 
in  the  ratio  of  two  given  straight  lines,    (y.  28.) 

PROPOSITION  XVli— PROBLEM.X)^ 


Q/n 


29.  To  construct  a  polygon  similar  to  a  given  polygon  P  and 
equivalent  to  a  given  polygon  Q. 

Find  M  and  iV,   the  sides  of 

squares  respectively  equal  in  area 

to  P  and  0  (23  and  24). 

I      ^      1         I     ^    1 
Let  a  be  any  side  of  P,  and  find 

a  fourth  proportional  a'  to  Mj  i^T,  \     p'     | 

and  a;  upon  a',  as  a  homologous  ^ — -f — ' 

side  to  a,  construct  the  polygon  P' 

similar  to  P;   this  will  be  the  required  polygon.     For,  by 

construction, 

M_a 

jsr     a" 

therefore,  taking  the  letters  P,  Q,  and  P',  to  den0te  the  areas 
of  the  polygons, 

P^3P^a^ 

Q       W'        a'' ' 


BOOK   IV. 


145 


But,  the  polygons  P  and  P'  being   similar,  we   have,   by 
(Proposition  IX,), 

P'       a'^ ' 

and  comparing  these  equations,  we  have  P'  =  Q. 

Therefore  the  polygon  P'  is  similar  to  the  polygon  P  and 
equivalent  to  the  polygon  Q,  as  required. 

EXERCISE. 


Problem. — To  construct  a  polygon  similar  to  a  given  polygon^ 
and  whose  area  shall  he  in  a  given  rhtio  to  that  of  the  given 
polygon,    (v.  28,  Exercise,  an^III.,  43.)    ^ 


■J 


# 


G       k      ^^^ 


i.^ 


IS 


D 


f 


EXERCISES  ON  BOOK  IV. 


THEOREMS. 


^r  ^       1.  Two' triangles  are  equivalent  if  they  have  two  sides  of  the 

(^r'^^iiAM)  one  respectively  equal  to  two  sides  of  the  other,  and  the  included 

At  angle  of  the  one  equal  to  the  supplement  of  the  included  angle 

of  the  other. 

-— ^^Vz.  The  two  opposite  triangles  formed  by  joining  any  point  in 

the  interior  of  a  parallelogram  4o  its  four  vertices  are  together 

equivalent  to  one-half  the  parallelogram. 

3.  The  triangle  formed  by  joining  the  middle  point  of  one  of 
the  non-parallel  sides  of  a  trapezoid  to  the  extremities  of  the 
opposite  side  is  equivalent  to  one-half  the  trapezoid,  {v,  I.,  Ex- 
ercise 24.) 

4.  The  figure  formed  by  joining  consecutively  the  four  middle 
points  of  the  sides  of  any  quadrilateral  is  equivalent  to  one-half 
the  quadrilateral,    (v.  I.,  Exercise  32.) 

„  V.  6.  If  in  a  rectangle  ABCD  we  draw 
the  diagonal  AC^  inscribe  a  circle  in  the 
triangle  ABC^  and  from  its  centre  draw 
OE  and  Oi^  perpendicular  to  AD  and 
DC  respectively,  the  rectangle  OD  will 
be  equivalent  to  one-half  the  rectangle 
ABCD. 

C  6.  The  area  of  a  triangle  is  equal  to 
one-half  the  product  of  its  perimeter  by 
the  radius  of  the  inscribed  circle. 


^ 


7.  The  area  of  a  rhonabus  is  one-half  the  product  of  the  diag- 
onals. 


.\,w^j'«' 


BOOK  IV.  147 

-  '"  8.  The  straight  line  joining  the  middle  points  of  the  parallel 
sides  of  a  trapezoid  divides  it  into  two  equivalent  figures. 

(4:'j  ^.  Any  line  drawn  through  the  point  of  intersection  of  the  diag- 
onals of  a  parallelogram  divides  it  into  two  equal  quadrilaterals. 

10.  In  an  isosceles  right  triangle  either  leg  is  a  mean  propor- 
tional between  the  hypotenuse  and  the  perpendicular  upon  it 
from  the  vertex  of  the  right  angle. 

11.  If  two  triangles  have  an  angle  in  common,  and  have  equal 
areas,  the  sides  about  the  equal  angles  are  reciprocally  propor- 

V  tional. 
.  .Ivt  \'^'  The  perimeter  of  a  triangle  is  to  a  side  as  the  perpendicular 
fronythe  opposite  vertex  is  to  the  radius  of  the  inscribed  circle. 
! :      (v.yExercise  6.) 

f    yi3.  Two  quadrilaterals  are  equivalent  when  the  diagonals  of  , 
V     one  are  respectively  equal  and  parallel  to  the  diagonals  of  the  I 
^o^her. 
^  14.  The  sum  of  the  perpendiculars  from  any  point  within  an 

■  ♦     equilateral  convex  polygon  upon  the  sides  is  constant. 

y^  Suggestion.  Join  the  point  with  the  vertices  of  the  polygon. 
^  15.  The  lines  joining  two  opposite  vertices  of  a  parallelogram 

^      with  the  middle  points  of  the  sides  form  a  parallelogram  whose 
^^^area  is  one-third  the  a^ea  of  the  given  parallelogram. 
A  ^.^    16.  The  sum  of  the  squares  on  the  segments  of  two  j)erpendic- 
ular  chords  in  a  circle  is  equivalent  to  the  square  on  the  diameter. 
17.  Let  ABC  be  any  triangle,  and 
"^     upon  the  sides  AB,  AC,  construct 
parallelograms  AD,  AF,  of  any  mag- 
nitude or  form.    Let  their  exterior 
*^     sides  DE,  FG,  meet  in  M;  join  MA, 
^   and  upon  BC  construct  a  parallelo- 
gram BK,  whose  side  BH  is  equal 
and  parallel  to  MA,    Then  the  par- 
allelogram BK  is  equivalent  to  the 
sum  of  the  parallelograms  AD  and 
AF,    {v.  Proposition  I.) 
From  this  deduce  the  Pythagorean  Theorem. 
«y^^  V     18.  Prove,  geometrically,  that  th^  square  described  upon  the 
sum  of  two  straight  lines  is  equivalent  to  the  sum  of  the  squares 
described  on  the  two  lines  plits  twice  their  rectangle. 

Note.  By  the  "rectangle  of  two  lines"  is  here  meant  the  rec- 
tangle  of  which  the  two  lines  are  the  adjacent  sides. 
QV'      19.  Prove,  geometrically,  that  the  square  described  upon  the 
difference  of  two  straight  lines  is  equivalent  to  the  sum  of  the 
squares  described  on  the  two  lines  minus  twice  their  rectangle. 


ELEMENTS   OP   GEOMETRY. 


V 

20.  Prove,  geometrically,  that  the  rectangle  of  the  sum  and  the 

difference  of  two  straight  lines  is  equivalent  to  the  difference  of 

the  squares  of  those  lines. 


PROBLEMS. 


.4 


V 


u 


21.  To  construct  a  triangle,  given  its  angles  and  its  area  (eq 
to  that  of  a  given  square). 

Suggestion.  Construct  any  triangle  having  the  given  angles. 
The  problem  then  reduces  to  (29). 

22.  Given  any  triangle,  to  construct  an  isosceles  triangle  of  the 
same  area,  whose  vertical  angle  is  an  angle  of  the  given  triangle. 
{v.  19,  Exercise.) 

23.  Given  any  triangle,  to  construct  an  equilateral  triangle  of 
he  same  area.    {v.  Exercise  21.) 

24.  Bisect  a  given  triangle  by  a  parallel  to  one  of  its  sides,  (v. 
Proposition  VIII.  and  28.) 

25.  Bisect  a  triangle  by  a  straight  line  drawn  through  a  given 
point  in  one  of  its  sides,    (v.  19,  Exercise.) 

26.  Inscribe  a  rectangle  of  a  given  area  in  a  given  circle. 
Suggestion.  Draw  a  diagonal  of  the  rectangle.    The  problem 

can  then  be  reduced  to  inscribing  in  the  given  circle  a  right 

;riangle  of  given  area. 
27.  Given  three  points,  A,  B^  and  (7,  to  find  a  fourth  point  P, 
such  that  the  areas  of  the  triangles  APB,  APC,  BPG,  shall  be 
equal.    (Four  solutions.)    (v.  III.,  Exercise  19.) 


h 


\ 


;^*(..«' 


'  BOOK  T. 

REGULAR  POLYGONS.     MEASUREMENT  ®P  THE 

CIRCLE. 

1.  Definition.  A  regular  polygon  is  a  polygon  which  is  at 
once  equilateral  and  equiangular. 

The  equilateral  triangle  and  the  square  are  simple  exam- 
ples of  regular  polygons.  The  following  theorem  establishes 
the  possibility  of  regular  polygons  of  any  number  of  sides. 


PROPOSITION  I.— THEOREM. 

2.  If  the  circumference  of  a  circle  be  divided  into  any  number 
of  equal  parts,  the  chords  joining  the  successive  points  of  division 
form  a  regular  polygon  inscribed  in  the  circle  ;  and  the  tangents 
drawn  at  the  points  of  division  form  a  regular  polygon  circum^ 
scribed  about  the  circle. 

Let  the  circumference  be  divided  into 
the  equal  arcs  AB,  BC,  GD^  etc. ;  then, 
1st,  drawing  the  chords  AB,  BC,  CD, 
etc.,  ABGD,  etc.,  is  a  regular  inscribed 
polygon.  For  its  sides  are  equal,  being 
chords  of  equal  arcs ;  and  its  angles  are 
equal,  being  inscribed  in  equal  segments. 

2d.  Drawing  tangents  at  A,  B,  C,  etc.,  the  polygon  GHK^ 
etc.,  is  a  regular  circumscribed  polygon.  For,  in  the  triangles 
AGB,  BHC,  CKD,  etc.,  we  have  AB  =  BC  =  CD,  etc.,  and 
the  angles  GAB,  GBA,  HBC,  HOB,  etc.,  are  equal,  since  each 

13*  149 


150 


ELEMENTS   OF   GEOMETRY. 


is  formed  by  a  tangent  and  chord  and  is  measured  by  half 
of  one  of  the  equal  parts  of  the  circum- 
ference (II.,  Proposition  XY.)  ;  therefore 
these  triangles  are  all  isosceles  and  equal 
to  each  other.  Hence  we  have  the  an- 
gles a  =  H=  K,  etc.,  and  AG  =  GB 
=.BH  =  HG  =  CK,  etc.,  from  which, 
by  the  addition  of  equals,  it  follows  that 
GH  =  HK,  etc. 

3.  Corollary  I.  If  the  vertices  of  a  regular  inscribed  poly- 
gon are  joined  with  the  middle  points  of  the  arcs  subtended  by  the 
sides  of  the  polygon,  the  joining  lines  will  form  a  regular  inscribed 
polygon  of  double  the  number  of  sides. 

4.  Corollary  II.  If  at  the  middle  points  of  the  arcs  joining 
adjacent  points  of  contact  of  the  sides  of  a  regular  circumscribed 
polygon  tangents  are  drawn,  a  regular  circumscribed  polygon  of 
double  the  number  of  sides  will  be  formed. 

5.  Scholium.  It  is  evident  that  the  area  of  an  inscribed 
polygon  is  less  than  that  of  the  inscribed  polygon  of  double 
the  number  of  sides ;  and  the  area  of  a  circumscribed  polygon 
is  greater  than  that  of  the  circumscribed  polygon  of  double 
the  number  of  sides. 

EXERCISE. 

Theorem. — If  a  regular  polygon  is  inscribed  in  a  circle,  the 
tangents  drawn  at  the  middle  points  of 
the  arcs  subtended  by  the  sides  of  the 
inscribed  polygon  form  a  circumscribed 
regular  polygon,  whose  sides  are  par- 
allel to  the  sides  of  the  inscribed  poly- 
gon, and  whose  vertices  lie  on  the  radii 
drawn  to  the  vertices  of  the  inscribed 
polygon. 


c 


A'{ 


""~--^A 

e/7  \ 

/\ 

V 

\f 

V"^'-^ 

^^\/ 

u 


BOOK   V.  151 


PROPOSITION  II.— THEOREM. 

6.  A  circle  may  be  circumscribed  about  any  regular  polygon ; 
and  a  circle  may  also  be  inscribed  in  it. 

Let  ABCD ...  be  a  regular  polygon. 
Through  A'  and  B\  the  middle  points 
of  AB  and  BG^  draw  perpendiculars, 
and  connect  0,  their  point  of  intersec- 
tion, with  all  the  vertices  of  the  poly- 
gon and  with  the  middle  points  of  all 
the  sides. 

The  triangles  OA'B  and  OB'B  are  equal,  by  I.,  Proposition 
X.  OB'B  and  OB'G  are  equal,  by  I.,  Proposition  YI.  The 
angle  OBB'  is  one-half  of  ABG ;  .  • .  005'  is  one-half  of  the 
equal  angle  BGD.  Hence  the  triangles  OB'G  and  OGG'  are 
equal,  by  I.,  Proposition  YI.  By  continuing  this  process  we 
may  prove  all  the  small  triangles  equal.  0,  then,  is  equidis- 
tant from  all  the  vertices,  and  therefore  with  0  as  a  centre  a 
circle  may  be  circumscribed  about  the  polygon.  0  is  also 
equidistant  from  all  the  sides,  and  therefore  with  0  as  a 
centre  a  circle  may  be  inscribed  in  the  polygon. 

7.  Definitions.  The  centre  of  a  regular  polygon  is  the  common 
centre,  0,  of  the  circumscribed  and  in- 
scribed circles. 

The  radius  of  a  regular  polygon  is 
the  radius,  OJ.,  of  the  circumscribed 
circle. 

The  apothem  is  the  radius,  O-ff,  of 
the  inscribed  circle. 

The  angle  at  the  centre  is  the  angle,  AOB,  formed  by  radii 
drawn  to  the  extremities  of  any  side. 


^o 

A 

vv 

^E 

152 


ELEMENTS   OF   GEOMETRY. 


8.  The  angle  at  the  centre  is  equal  to  four  right  angles 
divided  by  the  number  of  sides  of  the 

polygon. 

9.  Since  the  angle  ABC  is  equal  to 
twice  ABO,  or  to  ABO  +  BAO,  it  fol- 
lows that  the  angle  ABC  of  the  poly- 
gon is  the  supplement  of  the  angle  at 
the  centre. 


PROPOSITION  in.— THEOREM. 

10.  Begular  'polygons  of  the  same  number  of  sides  are  similar, 

Ijet  ABCI)JE,A'B'C'iyE\ 
be  regular  polygons  of  the 
same  number  of  sides ;  then 
they  are  similar. 

For,  1st,  they  ape  mutu- 
ally equiangular,  since  the 
magnitude  of  an  angle  of 

either  polygon  depends  only  on  the  number  of  the  sides  (8 
and  9),  which  is  the  same  in  both. 

2d.  The  homologous  sides  are  proportional,  since  the  ratio 
AB  :  A'B'  is  the  same  as  the  ratio  BG :  B'C,  or  CD  :  CD', 
etc. 

Therefore  the  polygons  fulfil  the  two  conditions  of  simi- 
larity. 

11.  Corollary.  The  perimeters  of  regular  polygons  of  the 
same  number  of  sides  are  to  each  other  as  the  radii  of  the  cir- 
cumscribed circles,  or  as  the  radii  of  the  inscribed  circles ;  and 
their  areas  are  to  each  other  as  the  squares  of  these  radii,  (y, 
III.,  Proposition  YIII.,  and  lY.,  Proposition  IX.) 


BOOK    V.  153 

PROPOSITION  IV.— THEOREM. 

12.  The  area  of  a  regular  polygon  is  equal  to  half  the  product 
of  its  perimeter  and  apothem. 

For  straight  lines  drawn  from  the  centre  to  the  vertices 
of  the  polygon  divide  it  into  equal  triangles  whose  bases  are 
the  sides  of  the  polygon  and  whose  common  altitude  is  the 
apothem.  The  area  of  one  of  these  triangles  is  equal  to 
half  the  product  of  its  base  and  altitude ;  therefore  the  sum 
of  their  areas,  or  the  area  of  the  polygon,  is  half  the  product 
of  the  sum  of  the  bases  by  the  common  altitude;  that  is, 
half  the  product  of  the  perimeter  and  apothem. 

EXERCISE. 

Theorem. — The  area  of  any  polygon  circumscribed  about  a 
circle  is  half  the  product  of  its  perimeter  by  the  radius  of  the 
circle. 

PROPOSITION  v.— THEOREM. 

13.  An  arc  of  a  circle  is  less  than  any  line  which  envelops  it 
and  has  the  same  extremities. 

Let  AKB  be  an  arc  of  a  circle,  AB  its 
chord ;  and  let  ALB,  AMB,  etc.,  be  any 
lines  enveloping  it  and  terminating  at  A 
and  B. 

Of  all  the  lines  AKB,  ALB,  AMB,  etc., 
which  can  be  drawn  (each  including  between  itself  and  the 
chord  AB  the  segment,  or  area,  AKB),  there  must  be  at  least 
one  minimum  or  shortest  line,  since  all  the  lines  are  obviously 
not  equal.  Now,  no  one  of  the  lines  ALB,  AMB,  etc.,  envel- 
oping  AKB,  can  be  such  a  minimum  ;  for,  drawing  a  tangent 
CKL  to  the  arc  AKB,  the  ImeAGJCLB  is  less  than  ACLDB ; 


154  ELEMENTS  OF  GEOMETRY. 

therefore  ALB  is  not  the  minimum ;  and  in  the  same  way  it 
is  shown  that  no  other  enveloping  line  can 
be  the  minimum.     Therefore  the  arc  AKB 
is  the  minimum. 

14.  Corollary.  The  circumference  of  a 
circle  is  less  than  the  perimeter  of  any  poly- 
gon circumscribed  about  it. 

15.  Scholium.  The  demonstration  is  applicable  when  AKB 
is  any  convex  curve  whatever. 


PKOPOSITION  VI.— THEOREM. 

16.  If  the  number  of  sides  of  a  regular  polygon  inscribed  in  a 
circle  be  increased  indefinitely,  the  apothem  of  the  polygon  will 
approach  the  radius  of  the  circle  as  its  limit. 

Let  AB  be  a  side  of  a  regular  polygon  in- 
scribed in  the  circle  whose  radius  is  OA; 
and  let  OD  be  its  apothem. 

Whatever  the  number  of  sides  of  the  poly- 
gon OB  <  OA,   by   I.,   Proposition   XYII. 
OA  <  AD  +  OD  (I.,  Axiom  I.)  •  ..  OA  —  OD  <,  AD,  and 
consequently  OA  —  OD  <^  AB. 

The  perimeter  of  the  polygon  is  manifestly  less  than  the 
circumference  of  the  circle.  If  n  is  the  number  of  sides  of 
the  polygon,  AB  is  less  than  one-nth  of  the  circumference. 
Therefore,  by  taking  a  sufficiently  great  value  of  n,  we  can 
make  AB,  and  consequently  OA  —  OD,  as  small  as  we  please. 

Since  OA  —  OD  can  be  made  as  small  as  we  please  by 
increasing  the  number  of  sides  of  the  polygon,  but  cannot  be 
made  absolutely  zero,  OA  is  the  limit  of  OD,  as  the  number 
of  sides  of  the  polygon  is  indefinitely  increased  (II.,  39). 


BOOK    V. 


155 


PROPOSITION  VII.— THEOREM. 

17.  The  circumference  of  a  circle  is  the  limit  which  the  perim- 
eters of  regular  inscribed  and  circumscribed  polygons  approach 
when  the  number  of  their  sides  is  increased  indefinitely ;  and  the 
area  of  the  circle  is  the  limit  of  the  areas  of  these  polygons. 

Let  AB  and  CD  be  sides  of  a  regular  in- 
scribed and  a  similar  circumscribed  polygon 
(v.  Proposition  I.,  Exercise)  ;  let  r  denote  the 
apothem  OE,  R  the  radius  OF^  p  the  perim- 
eter of  the  inscribed  polygon,  P  the  perim- 
eter of  the  circumscribed  polygon.  Then  we 
have  (Proposition  III.,  Corollary) 


whence,  by  division  (III.,  10), 
P  —  p  __R  —  r 


R 


or  P 


Now,  we  have  seen  in  Proposition  YI.  that  by  increasing 

the  number  of  sides  of  the  polygons  the  diiference  R  —  r  may 

P 

be  decreased  at  pleasure ;  consequently,  since  — -  does  not  in- 

R 

P 

crease,  — -  X  (-^  —  r),  or  P  —  p,  may  be  decreased  at  pleasure. 
R 

But  P  being  always  greater,  and  p  always  less,  than  the  cir- 
cumference of  the  circle,  the  difference  between  this  circum- 
ference and  either  P  or  ^  is  less  than  the  difference  P  -~  p^ 
and  consequently  may  be  made  as  small  as  we  please  by 
increasing  the  number  of  sides  of  the  polygons,  and  sinco  it 
obviously  cannot  be  made  absolutely  zero,  the  circumference 
is  the  common  limit  of  P  and  p,  as  the  number  of  sides  of 
the  polygons  is  indefinitely  increased. 


IjiisaL. 


156 


ELEMENTS   OF   GEOMETRY. 


Again,  let  s  and  S  denote  the  areas  of  two  similar  inscribed 
and  circumscribed  polygons.  The  difference 
between  the  triangles  COD  and  A  OB  is  the 
trapezoid  CABD,  the  measure  of  which  is 
^ (CD  +  AB)  X  ^^;  therefore  the  difference 
between  the  areas  of  the  polygons  is 


C           F 

D 

±\^ 

^\/« 

N 

^ 

^ 

consequently, 


S-s<PXiB-r). 


Now,  by  increasing  the  number  of  sides  of  the  polygons  the 
quantity  P  X  {^  —  0)  ^^^  consequently  also  S  —  5,  may  be 
decreased  at  pleasure.  But  S  being  always  greater,  and  5 
always  less,  than  the  area  of  the  circle,  the  difference  between 
the  area  of  the  circle  and  either  S  ov  s  is  less  than  the  differ- 
ence S  —  5,  and  consequently  may  also  be  made  as  small  as 
we  please  by  increasing  the  number  of  sides  of  the  polygons, 
and  since  it  obviously  cannot  be  made  absolutely  zero,  the 
area  of  the  circle  is  the  common  limit  of  S  and  5,  as  the 
number  of  sides  of  the  polygons  is  indefinitely  increased. 


PROPOSITION  VIII.— THEOREM. 

18.  The  circumferences  of  two  circles  are  to  each  other  as  their 
radii,  and  their  areas  are  to  each  other  as  the  squares  of  their 
radii. 

Let  B  and  B'  be  the  radii 
of  the  circles,  C  and  0'  their 
circumferences,  S  and  S'  their 
areas. 

Inscribe  in  the  two  circles 
similar  regular  polygons  of  any 
arbitrarily  chosen  number,  n,  of  sides ;  let  P  and  P'  denote 


BOOK   V.  157 

the  perimeters,  A  and  A'  the  areas,  of  these  polygons ;  then, 
the  polygons  being  similar,  we  have  (Proposition  III.,  Corol- 
lary) 

P  ^  ^  A^  R^ 

P'       J2"        A'       B'^' 


no  matter  what  the  value  of  n.  ^ 


'  Up 

As  we  change  n,  P  and  --  X  P'  change,  but  remain  always 

equal  to  each  other. 

As  n  is  indefinitely  increased,  P  approaches  the  limit  0,  and 

7?  7? 

^^  X  P'  approaches  the  limit  ^^  X  C".     Therefore,  by  II., 

Theorem  of  Limits,  these  limits  are  equal,  and  we  have 

c  =  |xo', 

or 

^  =  ?L 

C      B!' 

In  the  same  way  we  may  prove 

K  =  ^ 
S'       It'' 

19.  Corollary  I.  The  circumferences  of  circles  are  to  each 
other  as  their  diameters,  and  their  areas  are  to  each  other  as  the 
squares  of  their  diameters. 

Suggestion.  D  =  2P,  if  D  is  the  diameter  and  R  the  ra- 
dius. 

14 


158  ELEMENTS  OF  GEOMETRY. 

20.  Corollary  II.  The  ratio  of  the  circumference  of  a  circle 

to  its  diameter  is  constant ;  that  is,  it  is  the  same  for  all  circles. 

O        2R 
For,  from  —  =  — — ,  we  have  at  once 

2B       2B'' 

This  constant  ratio  is  usually  denoted  by  tt,  so  that  for  any 
circle  whose  diameter  is  2R  and  circumference  G  we  have 

2R         ' 

21.  Scholium.  The  ratio  r  is  incommensurable  (as  can  be 
proved  by  the  higher  mathematics),  and  can  therefore  bo 
expressed  in  numbers  only  approximately.  The  letter  tt, 
however,  is  used  to  symbolize  its  exact  value. 

22.  Definitions.  Similar  arcs  are 
those  which  subtend  equal  angles 
at  the  centres  of  the  circles  to 
which  they  belong. 

Similar  sectors  are  sectors  whose  / 

bounding  radii  include  equal  angles.     ^     /^'^^^ .     Z^^-^^Ce^ 


EXERCISE. 

Theorem. — Similar  arcs  are  to  each  other  as  their  radii,  and 
similar  sectors  are  to  each  other  as  the  squares  of  their  radii. 

Suggestion.  The  arcs  are  like  parts  of  their  respective  cir- 
cumferences, and  the  sectors  like  parts  of  their  circles. 


BOOK  V.  159 

PROPOSITION  IX.— THEOREM. 

23.  The  area  of  a  circle  is  equal  to  half  the  product  of  its 
circumference  by  its  radius. 

Let  the  area  of  any  regular  polygon  cir- 
cumscribed about  the  circle  be  denoted  by 
A,  its  perimeter  by  P,  and  its  apothem, 
which  is  equal  to  the  radius  of  the  circle,  by 
B ;  and  let  S  be  the  area  and  C  the  circum- 
ference of  the  circle.     Then  A=  ^F  y^  R 
(Proposition  IY.)j  ^^  matter  what  the  number  of  sides  of 
the  polygon.     If  we   change   the   number  of  sides  of  the 
polygon,  A  and  i  P  X  -K  change,  but  remain  always  equal  to 
each  other. 

As  the  number  of  sides  is  indefinitely  increased,  A  ap- 
proaches the  limit  S^  and  ^ P  X  ^  the  limit  ^C  X  R-  There- 
fore, by  II.,  Theorem  of  Limits,  these  limits  are  equal,  and 
we  have 

>Sf=^(7xP.  [1] 

24.  Corollary.  The  area  of  a  circle  is  equal  to  the  square 
of  its  radius  multiplied  by  the  constant  number  tz. 

Suggestion.  If  we  substitute  for  C  in  [1]  its  value  27zR 
(20),  we  have 


EXERCISE. 

Theorem. — The  area  of  a  sector  is  equal  to  half  the  product 
of  its  arc  by  the  radius. 

Suggestion.  Compare  the  sector  with  the  whole  circle. 


160 


ELEMENTS  OF  GEOMETRY. 


PEOBLEMS  OF  COIS^STEUCTIOK  AND  COMPUTA- 

TIOK 

PKOPOSITION  X.— PROBLEM. 

25.  To  inscribe  a  square  in  a  given  circle. 

B 

Draw  any  two  diameters  A  C,  BJ),  per- 
pendicular to  each  other,  and  join  their 
extremities  by  the  chords  ABj  BC,  CD, 
DA;  then  ABCD  is  an  inscribed  square. 


26.  Corollary.  To  circumscribe  a  square  about  a  circle, 
draw  tangents  at  the  extremities  of  two  perpendicular  diam- 
eters AC,  BD. 

27.  Scholium.  In  the  right  triangle  ABO  we  have  AB'  = 
'UA'  +  'UB'=2'UA\  whence  AB  =OA.y^,  by  which  the 
side  of  the  inscribed  square  can  be  computed,  the  radius 
being  given. 

PROPOSITION  XI.— PROBLEM. 

28.  To  inscribe  a  regular  hexagon  in  a  given  circle. 

Suppose  the  problem  solved,  and  let 
ABCDEF  be  a  regular  inscribed  hexa- 
gon. 

Draw  the  radii  OA  and  OB.  The 
angle  A  OB  is  measured  by  \  of  the  cir- 
cumference, and  therefore  contains  60°. 

OAB  and  OB  A  are  therefore  together  equal  to  180°  —  60°, 
or  120° ;  and,  since  they  are  equal,  each  is  60°,  and  the  tri- 
angle OAB  is  equilateral,  and  th^gfor^the  side  of  the  in- 
scribed regular  hexagon  is  equal  to  thfe  radius  of  the  circle. 


BOOK   V. 


161 


Consequently,  to  inscribe   a  regular  hexagon,  apply  the 
radius  to  the  circle  six  times  as  a  chord. 

29.  Corollary.  To  inscribe  an  equilateral  triangle^  join  the 
alternate  vertices  of  the  regular  hexagon. 

30.  Scholium.  Since  OB  bisects  the  arc  ABC,  it  bisects  the 

chord  AG  Bit  right  angles ;  and  since  in  the  isosceles  triangle 

A  OB,  AH  is  perpendicular  to  the  base,  it  bisects  the  base, 

and 

OII=WB  =  WA; 

that  is,  the  apothem  of  an  inscribed  regular  triangle  is  equal 
to  one-half  the  radius. 

In  the  right  triangle  AHO,  AH^  =  WC  —  OTT'  =  TJA^  — 
{lOAy=^lUA\2^n^ 

whence  AC  =^  OA-^/S,  by  which  the  side  of  the  inscribed  tri- 
angle can  be  computed  from  the  radius. 

The  apothem  of  the  regular  inscribed  hexagon  is  equal  to 

OA 

2 


AS 


1/5- 


PROPOSITION  XII.— PROBLEM. 

31.   To  inscribe  a  regular  decagon  in  a  given  circle. 

Suppose  the  problem  solved,  and  let 
ABC .,. . .  Jv  be  a  regular  inscribed  deca- 
gon. 

Join  AFj  BGr ;  since  each  of  these 
lines  bisects  the  circumference,  they  are 
diameters  and  intersect  in  the  centre  0. 
Draw  BK  intersecting  OA  in  M. 

The  angle  AMB  is  measured  by  half 
the  sum  of  the  arcs  KF  and  AB  (II.,  Proposition  XYI.), — 

I  14* 


t'X^ 


ELEMENTS  OF   GEOMETRY. 


that  is,  )by  two  divi^ns  of  the  circumference ;  the  inscribed 
angle^ikHJ^  is  measured  by  half  the  arc 
BF^ — that  is,   also,  by  two  divisions; 
therefore  AMB  is  an  isosceles  triangle, 
and  MB  =  AB. 

Again,  the  inscribed  angle  MBO  is 
measured  by  half  the  arc  KG, — ^that  is, 
by  one  division ;  and  the  angle  MOB  at 
the  centre  has  the  safne  measure ;  there- 
fore 0MB  is  an  isosceles  triangle,  and  OM  =  MB  =  AB. 

The  inscribed  angle  MBA,  being  measured  by  half  the  arc 
AK, — that  is,  by  one  division, — is  equal  to  the  angle  AOB. 
Therefore  the  isosceles  triangles  AMB  and  A  OB  are  mutually 
equiangular  and  similar,  and  give  the  proportion 


whence 


OA'.AB^AB:  AM; 
OA  XAM  =  A^=(m\- 


that  is,  the  radius  OA  is  divided  in  extreme  and  mean  ratio 
at  Jf  (III.,  41)  ;  and  the  greater  segment  OJf  is  equal  to  the 
side  AB  of  the  inscribed  regular  decagon. 

Consequently,  to  inscribe  a  regular  decagon,  divide  the 
radius  in  extreme  and  mean  ratio  (III.,  42),  and  apply  the 
greater  segment  ten  times  as  a  chord. 

o 


32.  Corollary.  To  inscribe  a  regular 
pentagon  J  join  the  alternate  vertices  of 
the  regular  inscribed  decagon. 


BOOK   V.  163 

PROPOSITION  XIII.— PROBLEM. 

33.  To  inscribe  a  regular  pentedecagon  in  a  given  circle. 
Suppose  AB  is  the  side  of  a  ■  f 

regular  inscribed  pentedecagon,      ^n*-- - -y^ 

or  that  the  arc  AB  is  ^  of  the  ^'^^===*==>.-^i^i;:i^^^'''^ 

circumference. 

JSTow,  the  fraction  J^.  ==  i.  _  -i^ ;  therefore  the  arc  AB  is  the 
difference  between  \  and  -^  of  the  circumference.  Hence, 
if  we  inscribe  the  chord  AG  equal  to  the  side  of  the  regular 
inscribed  hexagon,  and  then  CB  equal  to  that  of  the  regular 
inscribed  decagon,  the  chord  AB  will  be  the  side  of  the 
regular  inscribed  pentedecagon  required. 

34.  Scholium.  Any  regular  inscribed  polygon  being  given, 
a  regular  inscribed  polygon  of  double  the  number  of  sides 
can  be  formed  by  bisecting  the  arcs  subtended  by  its  sides 
and  drawing  the  chords  of  the  semi-arcs  (Proposition  I.,  Cor- 
ollary I.).  Also,  any  regular  inscribed  polygon  being  given^ 
a  regular  circumscribed  polygon  of  the  same  number  of  sides 
can  be  formed  (Proposition  I.).  Therefore,  by  means  of  the 
inscribed  square,  we  can  inscribe  and  circumscribe,  succes- 
sively, regular  polygons  of  8,  16,  32,  etc.,  sides ;  by  means  of 
the  hexagon,  those  of  12,  24,  48,  etc.,  sides ;  by  means  of  the 
decagon,  those  of  20,  40,  80,  etc.,  sides ;  and,  finally,  by  means 
of  the  pentedecagon,  those  of  30,  60,  120,  etc.,  sides. 

Until  the  beginning  of  the  present  century,  it  was  sup- 
posed that  these  were  the  only  polygons  that  could  be  con- 
structed by  elementary  geometry ;  that  is,  by  the  use  of  the 
straight  line  and  circle  only.  Gaxjss,  however,  in  his  Disqui- 
sitiones  Arithmeticce^  Lipsise,  1801,  proved  that  it  is  possible, 
by  the  use  of  the  straight  line  and  circle  only,  to  construct 
regular  polygons  of  17  sides,  of  257  sides,  and  in  general  of 


>)'' nxXcj__^M3  -u^^JM" 


0              F           E 

G              D 

\  ^^^^P 

"-^4 

\    \    1 

\\i  , 

0 

/ 

164  - 


any  number  oF'sT^Fg  which  can  be^  expressed  by  2"  +  !>  ^ 
being  an  integer,  provided  that  2"  +  1  is  a  prime  number. 

I^ROPOSITION  XIV.— PROBLEM. 

35.  Given  the  perimeters  of  a  regular  inscribed  and  a  similar 
•  circumscribed  polygon,  to  compute  the  perimeters  of  the  regular 
inscribed  and  circumscribed  polygons  of  double  the  number  of 
sides. 

Let  AB  he  a  side  of  the  given 
inscribed  polygon,  and  CD  a  side 
of  the  similar  circumscribed  poly- 
gon, tangent  to  the  arc  AB  Sit  its 
middle  point  B.  Join  AE,  and  at 
A  and  B  draw  the  tangents  AF 
and  BG ;  then  AE  is  a  side  of  the 

regular  inscribed  polygon  of  double  the  number  of  sides,  and 
FG  is  a  side  of  the  circumscribed  polygon  of  double  the 
number  of  sides. 

Denote  the  perimeters  of  the  given  inscribed  and  circum- 
scribed polygons  by  p  and  P,  respectively ;  and  the  perimeters 
of  the  required  inscribed  and  circumscribed  polygons  of 
double  the  number  of  sides  by  y  and  P',  respectively. 

Since  0(7  is  the  radius  of  the  circle  circumscribed  about 
the  polygon  whose  perimeter  is  P,  we  have  (Proposition  III., 
Corollary) 

?=oc^^qc 

p        OA       OE' 
and  since  OF  bisects  the  angle  COE,  we  have  (III.,  15,  Ex- 
ercise) 

OC^CF, 
OE      FE' 
therefore 

P^CF 
p       FE' 


BOOK   V.  165 

whence,  by  composition, 

P  +  p  ^  CF-\-FE^  CE 
•       2p  2FE  Fa' 

I^ow,  FG  is  a  side  of  the  polygon  whose  perimeter  is  P',  and 
is  contained  as  many  times  in  F'  as  GE  is  contained  in  P; 
hence  (III.,  9) 


CE       P 
FG       F" 

and  therefore 

P+P       P^ 

whence 

2p           F' 

F  +  p 

m 

Again,  the  right  triangles  AES  and  EFW  are  similar,  since 
their  acute  angles  EAS  and  FEJ}^  are  equal,  and  give 

AH^EN 
AE       EF 

Since  AH  and  AE  are  contained  the  same  number  of  times 
in  p  and  p\  respectively,  we  have 

AH  ^p  , 
AE      ^' 

and  since  EN  and  EF  are  contained  the  same  number  of 
times  in  »'  and  P',  respectively,  we  have 

^^  EW^p^, 

EF      P" 

therefore  we  have 

p'=y^^lxrT\  ^2^ 

Therefore,  from  the  given  perimeters  p  and  P  we  compute 
P'  by  the  equation  [1],  and  then  with  p  and  P'  we  compute 
p'  by  the  equation  r2"|. 


whence 


166  ELEMENTS   OF   GEOMETRY. 


PROPOSITION  XV.— PROBLEM. 

36.  To  compute  the  ratio  of  the  circumference  of  a  circle  to  its 
diameter^  approximately. 

Method  op  Perimeters. — In  this  method,  we  take  the 
diameter  of  the  circle  as  given  and  compute  the  perimeters 
of  some  inscribed  and  a  similar  circumscribed  regular  poly- 
gon. We  then  compute  the  perimeters  of  inscribed  and  cir- 
cumscribed regular  polygons  of  double  the  number  of  sides, 
by  Proposition  XIY.  Taking  the  last-found  perimeters  as 
given,  we  compute  the  perimeters  of  polygons  of  double  the 
number  of  sides  by  the  same  method ;  and  so  on.  Each  com- 
putation gives  us,  of  course,  a  pair  of  values  between  which 
the  value  of  the  circumference  must  lie.  As  we  continue  the 
process,  these  values  will  come  nearer  and  nearer  to  the 
actual  value  of  the  circumference  (Proposition  YII.),  and  we 
may  thus  obtain  as  close  an  approximation  to  that  value  as 
we  please. 

Taking,  then,  the  diameter  of  the  circle  as  given  =  1,  let 
us  begin  by  inscribing  and  circumscribing  a  square.     The" 
perimeter  of  the  inscribed  square  =  4  X  ?  X  i/^=  2y^  (27) ; 
J  that  of  the  circumscribed  square  =  4 ;  therefore,  putting 

p  =  2^2  =t  2.8284271,        S  r     I    '^   ' 

}  we  find,  by  Proposition  X.,  for  the  perimeters  of  the  circum- 

scribed and  inscribed  regular  octagons. 


P'  =  ^P^^  =  3.3187085, 
p'  =  yp  xP'  =  3.0614675. 


BOOK  V.  167 

Then,  taking  these  as  given  quantities,  we  put 

P  =  3.3137085,  p  =  3.0614675, 
and  find  by  the  same  formulae  for  the  polygons  of  16  sides 

P'  ==  3.1825979,  /  =  3.1214452. 

Continuing  this  process,  the  results  will  be  found  as  in  the 
following 

TABLE.* 


Number 

Perimeter  of 

Perimeter  of 

of  sides. 

circumscribed  polygon. 

inscribed  polygon. 

4 

4.0000000 

2.8284271 

8 

3.3137085 

3.0614675 

16 

3.182.5979 

3.1214452 

32 

3.1517249 

3.1365485 

64 

3.1441184 

3.1403312 

128 

3.1422236 

3.1412773 

256 

3.1417504 

3.1415138 

512 

3.1416321 

3.1415729 

1024 

3.1416025 

3.1415877 

2048 

3.1415951 

3.1415914 

4096 

3.1415933 

3.1415923 

8192 

3.1415928 

3.1415926 

From  the  last  two  numbers  of  this  table  we  learn  that  the 
circumference  of  the  circle  whose  diameter  is  unity  is  less 
than  3.1415928  and  greater  than  3.1415926 ;  and  since,  when 
the  diameter  =  1,  we  have  (7  ==  tt  (20),  it  follows  that 

TT  =  3.1415927 

within  a  unit  of  the  seventh  decimal  place. 

*  The  computations  have  been  carried  out  with  ten  decimal  places  in 
order  to  insure  the  accuracy  of  the  seventh  place,  as  given  in  the  table. 


168  ELEMENTS  OF  GEOMETRY. 

37.  Scholium.  Archimedes  (bom  287  b.c.)  was  the  first  to 
assign  an  approximate  value  of  ;:.  By  a  method  similar  to 
that  above,  he  proved  that  its  value  is  between  3^  and  3^J, 
or,  in  decimals,  between  3.1428  and  3.1408;  he  therefore  as- 
signed its  value  correctly  within  a  unit  of  the  third  decimal 
place.  The  number  3^,  or  %^^  usually  cited  as  Archimedes' 
value  of  Tz  (although  it  is  but  one  of  the  two  limits  assigned 
by  him),  is  often  used  as  a  sufficient  approximation  in  rough 
computations. 

Metius  (a.d.  1640)  found  the  much  more  accurate  value 
1^,  which  correctly  represents  even  the  sixth  decimal  place. 
It  is  easily  remembered  by  observing  that  the  denominator 
and  numerator  written  consecutively,  thus,  113 1 355,  present 
the  first  three  odd  numbers  each  written  twice. 

More  recently,  the  value  has  been  found  to  a  very  great 
number  of  decimals,  by  the  aid  of  series  demonstrated  by 
the  Differential  Calculus.  Clausen  and  Dase,  of  Germany 
(about  A.D.  1846),  computing  independently  of  each  other, 
carried  out  the  value  to  two  hundred  decimal  places,  and 
their  results  agree  to  the  last  figure.  The  mutual  verifica- 
tion thus  obtained  stamps  their  results  as  thus  far  the  best 
established  value  to  the  two-hundredth  place.  (See  Schu- 
macher's Astronomische  Nachrichten,  No.  589.)  Other  com- 
puters have  carried  the  value  to  over  five  hundred  places, 
but  it  does  not  appear  that  their  results  have  been  verified. 

The  value  to  fifteen  decimal  places  is 

TT  =  3.141592653589793. 

For  the  greater  number  of  practical  applications,  the  value 
t:  =  3.1416  is  sufficiently  accurate.  *~— - 


(^ 


IP 


EXERCISES  ON  BOOK  V. 


THEOREMS. 


^1.  An  equilateral  polygon  inscribed  in  a  circle  is  regular. 

^  2.  An  equilateral  polygon  circumscribed  about  a  circle  is  reg- 
ular if  the  number  of  its  sides  is  odd. 

^  3.  An  equiangular  polygon  inscribed  in  a  circle  is  regular  if  the 
number  of  its  sides  is  odd. 

\  4.  An  equiangular  polygon  circumscribed  about  a  circle  is  reg- 
ular. 

\i  6.  The  area  of  the  regular  inscribed  triangle  is  one-half  the  area 
of  the  regular  inscribed  hexagon. 

/  6.  The  area  of  the  regular  inscribed  hexagon  is  three-fourths  of 
that  of  the  regular  circumscribed  hexagon. 

I  7.  The  area  of  the  regular  inscribed  hexagon  is  a  mean  propor- 
tional between  the  areas  of  the  inscribed  and  circumscribed  equi- 
lateral triangles. 

(L  \  8.  A  plane  surface  may  be  entirely  covered  (as  in  the  construc- 
tion of  a  pavement)  by  equal  regular  polygons  of  either  three, 
four,  or  six  sides. 

9.  A  plane  surface  may  be  entirely  covered  by  a  combination 
of  squares  and  regular  octagons  having  the  same  side,  or  by 
dodecagons  and  equilateral  triangles  having  the  same  side. 

10.  If  squares  be  described  on  the  sides  of  a  regular  hexagon, 
and  their  adjacent  external  vertices  be  joined,  a  regular  dodeca- 
gon will  be  formed. 

11.  The  diagonals  of  a  regular  pentagon  form  a  regular  pentagon. 

12.  The  diagonals  joining  alternate  vertices  of  a  regular  hexagon 
enclose  a  regular  hexagon  one-third  as  large  as  the  original  hex- 
agon. .  \ 


15  169 


k 


170 


ELEMENTS   OF   GEOMETRY. 


13.  The  area  of  the  regular  inscribed  octagon  is  equal  to  the 
product  of  the  side  of  the  inscribed  square  by  the  diameter. 

Suggestion.  A  quarter  of  the  octagon  is  the  sum  of  two  triangles 
having  as  a  common  base  the  side  of  the  inscribed  square,  and 
having  the  radius  as  the  sum  of  their  altitudes. 


14.  The  area  of  a  regular  inscribed 
dodecagon  is  equal  to  three  times  the 
square  of  the  radius. 


15.  Prove  the  correctness  of  the  following 
construction : 

If  AB  and  CD  are  two  perpendicular 
diameters  in  a  circle,  and  E  the  middle 
point  of  the  radius  0(7,  and  if  EFis  taken 
equal  to  EA^  then  OF  is  equal  to  the  side 
of  the  regular  inscribed  decagon,  and  AF 
is  equal  to  the  side  of  the  regular  inscribed 
pentagon,    {v.  III.,  42.) 


16.  From  any  point  within  a  regular  polygon  of  n  sides,  per- 
pendiculars are  drawn  to  the  several  sides ;  prove  that  the  sum 
of  thea^  perpendiculars  is  equal  to  n  times  the  apothem. 

Suggestion.  Join  the  point  with  the  vertices  of  the  polygon,  and 
obtain  an  expression  for  the  area  in  terms  of  the  perpendiculars  : 
then  see  Proposition  IV. 

17.  The  side  of  the  regular  inscribed  triangle  is  equal  to  the 
hypotenuse  of  a  right  triangle  of  which  the  sides  of  the  inscribed 
square  and  of  the  regular  inscribed  hexagon  are  the  sides,  {v. 
IV.,  Proposition  X.) 

18.  If  a  is  the  side  of  a  regular  decagon  inscribed  in  a  circle 
whose  radius  is  i?, 


u 


a  =  |  (1/5-1). 


B 


Suggestion.  By  (31),  -  = 

a       H 


a 


BOOK   V. 


171- 


19.  If  a  =  the  side  of  a  regular  polygon  inscribed  in  a  circle 
whose  radius  is  i?,  and  a'  =  the  side  of  the 
regular  inscribed  polygon  of  double  the 
number  of  sides,  then 


a^2  =r{2R  —  VU^ 
a 


0, 


Suggestion.  ABC  and  ADO  are  similar. 
Hence  ^  =  ^  and  ^UfVbut  37)^ 

=  {2RY  —  a'\ 


c 

7  a" 

\      ''' 


20.  If  a  =  the  side  of  a  regular  pentagon  inscribed  in  a  circle 
whose  radius  is  i?,  then 


R. 


21.  If  a  =  the  side  of  a  regular  octagon  inscribed  in  a  circle 
whose  radius  is  B^  then 


a  =  BV2  —  y± 


22.  If  a  =  the  side  of  a  regular  dodecagon  inscribed  in  a  circle 
whose  radius  is  i2,  then 


=ri2l/2-l/5. 


23.  The  side  of  the  regular  inscribed  pentagon  is  equal  to  the 
hypotenuse  of  a  right  triangle  whose  sides  are  the  radius  and  the 
side  of  the  regular  inscribed  decagon. 

24.  The  area  of  a  ring  bounded  by  two  concentric  circumfer- 
ences is  equal  to  the  area  of  a  circle  having  for  its  diameter  a 
chord  of  the  outer  circumference  tangent  to  the  inner  circumfer- 
ence. 


25.  If  on  the  legs  of  a  right  triangle, 
as  diameters,  semicircles  are  described 
external  to  the  triangle,  and  from  the 
whole  figure  a  semicircle  on  the  hypot- 
enuse is  subtracted,  the  remainder  is 
equivalent  to  the  given  triangle. 


I^*r%.r 


172 


ELEMENTS   OF   GEOMETRY. 


26.  If  on  the  two  segments  into  which  a  diameter  of  a  given 
circle  is  divided  by  any  point,  as  diameters, 
semi-circumferences  are  described  lying  on  op- 
posite sides  of  the  given  diameter,  the  sum  of 
their  lengths  is  equal  to  the  length  of  a  semi- 
circumference  of  the  given  circle,  and  a  line 
which  they  form  divides  the  circle  into  two 
parts  whose  areas  are  to  each  other  as  the  seg- 
ments of  the  given  diameter. 


27.  If  a  diameter  of  a  given  circle 
is  divided  into  n  equal  parts,  and 
through  each  point  of  division  a 
curved  line  of  the  sort  described  in 
the  last  problem  is  drawn,  these  lines 
will  divide  the  circle  into  n  equivar 
lent  parts. 


28.  If  a  circle  rolls  around  the 
circumference  of  a  circle  of  twice 
its  radius,  the  two  circles  being 
always  tangent  internally,  the 
locus  of  a  fixed  point  on  the  cir- 
cumference of  the  rolling  circle 
is  a  diameter  of  the  fixed  circle. 


^^vsIfUA-^ygasguareis  subdivided  into  n^ 
equal  squares,  nbSngany^^i^vm^number,  and 
in  each  of  these  smaller  squares  a  circle  is  in- 
scribed, the  sum  of  their  areas  is  equal  to  the 
area  of  the  circle  inscribed  in  the  original 
square. 


MISCELLANEOUS  EXERCISES 

PLANE    GEOMETRY. 


THEOREMS. 

1.  The  sum  of  the  three  straight  lines  drawn  from 
any  point  within  a  triangle  to  the  three  vertices  is 
less  than  the  sum  and  greater  than  the  half  sum 
of  the  three  sides  of  the  triangle. 


2.  If  one  of  the  acute  angles  of  a  right  triangle  is  double  the 
other,  the  hypotenuse  is  double  the  shortest  side. 


3.  If  from  any  point  within  an  equilateral 
triangle  perpendiculars  to  the  three  sides  are 
drawn,  the  sum  of  these  lines  is  constant,  and 
equal  to  the  perpendicular  from  any  vertex 
upon  the  opposite  side. 


4.  Lines  drawn  from  one  vertex  of  a 
parallelogram  to  the  middle  points  of  the 
opposite  sides  trisect  a  diagonal. 


6.  The  bisectors  of  the  angles  con- 
tained by  the  opposite  sides  (produced) 
of  an  inscribed  quadrilateral  intersect 
at  right  angles. 


15* 


174 


ELEMENTS   OF   GEOMETRY. 


6.  If  AOB  is  any  given  angle  at  the 
centre  of  a  circle,  and  if  BC  can  be 
drawn  meeting  AO  produced  in  C,  and 
the  circumference  in  Z>,  so  that  CD  shall 
be  equal  to  the  radius  of  the  circle,  then 
the  angle  C  will  be  equal  to  one-third  the 
angle  AOB. 

Note,  There  is  no  method  known  of  drawing  J5(7,  under  these 
conditions,  and  with  the  use  of  straight  lines  and  circles  only, 
AOB  being  any  given  angle  ;  so  that  the  trisection  of  an  angle^  in 
general,  is  a  problem  that  cannot  be  solved  by  elementary  geom- 
♦*try. 


7.  If  through  P,  one  of  the  points 
of  intersection  of  two  circumfer- 
ences, any  two  secants,  APB^  CPD^ 
are  drawn,  the  straight  lines,  AC, 
DB,  joining  the  extremities  of  the 
secants,  make  a  constant  angle  PJ, 
equal  to  the  angle  3fPN  formed  by 
the  tangents  at  P. 


8.  If  a  figure  is  moved  in  a  plane,  it  may  be  brought  from  one 
position  to  any  other  by  revolving  it  about  a  certain  fixed  point ; 
that  is,  by  causing  each  point  of  the  figure  to  move  in  the  circum- 
ference of  a  circle  whose  centre  is  the  fixed  point. 

9.  If  a  square  DEFO  is  inscribed  in  a  right  triangle  ABC,  so 
that  a  side  DE  coincides  with  the  hypotenuse  BC  (the  vertices  F 
and  O  being  in  the  sides  AC  and  AB),  then  the  side  DE  is  a 
mean  proportional  between  the  segments  BD  and  EC  of  the 
hypotenuse. 

10.  If  the  middle  points  of  the  sides  of  a  triangle  are  joined  by 
straight  lines,  the  medial  lines  of  the  triangle  thus  formed  are 
the  medial  line^  of  the  original  triangle,  and  the  perpendiculars 
from,  the  vertices  upon  the  opposite  sides  are  the  perpendiculars 
at  the  middle  poihts  of  the  sides  of  the  original  triangle. 

11.  If  O  is  the  centre  of  the  circle  circumscribed  about  a  triangle 
ABC,  and  P  is  the  intersection  of  the  perpendiculars  from  the 
angles  upon  the  opposite  sides,  the  perpendicular  from  O  upon 
the  side  JSC  is  equal  to  one-half  the  distance  AP. 


MISCELLANEOUS   EXERCISES. 


175 


12.  In  any  triangle,  the  centre  of  the  circumscribed  circle,  the 
intersection  of  the  medial  lines,  and  the  intersection  of  the  per- 
pendiculars from  the  angles  upon  the  opposite  sides,  are  in  the 
same  straight  line ;  and  the  distance  of  the  first  point  from  the 
second  is  one-half  the  distance  of  the  second  from  the  third. 

13.  If  two  circles  intersect  in  the  points  A  and  B,  and  through 
A  any  secant  CAD  is  drawn  terminated  by  the  circumferences  at 
C  and  i),  the  straight  lines  BC  and  BD  are  to  each  other  as  the 
diameters  of  the  circles. 

14.  If  through  the  middle  point  of  each  diagonal  of  any  quad- 
rilateral a  parallel  is  drawn  to  the  other  diagonal,  and  from  the 
intersection  of  these  parallels  straight  lines  are  drawn  to  the 
middle  points  of  the  four  sides,  these  straight  lines  divide  the 
quadrilateral  into  four  equivalent  parts. 


15.  If  three  straight  lines  Aa,  Bb,  Ce^ 
drawn  from  the  vertices  of  a  triangle  ABC 
to  the  opposite  sides,  pass  through  a  com- 
mon point  O  within  the  triangle,  then 


Oa  ,   Ob   ,0g  ^. 
Aa"^  Bb~^  Cc 


16.  If  from  any  point  O  within  a  tri- 
angle ABC  any  three  straight  lines,  Oa, 
Ob,  Oc,  are  drawn  to  the  three  sides,  and 
through  the  vertices  of  the  triangle  three 
straight  lines,  Aa^,  Bb\  Co',  are  drawn 
parallel  respectively  to  Oa,  Ob,  Oc,  then 

Oa   ^^\\    Q^  —I 
Aa^  T  Bb'  "^  Cc' 

17.  The  area  of  a  circle  is  a  mean  proportional  between  the 
areas  of  any  two  similar  polygons,  one  of  which  is  circumscribed 
about  the  circle  and  the  other  isoperimetrical  with  the  circle. 
{Galileo^ 8  Theorem.) 


18.  Two  diagonals  of  a  regular  pentagon,  not  drawn  from  a 
common  vertex,  divide  each  other  in  extreme  and  mean  ratio. 


176 


ELEMENTS   OF   GEOMETRY. 


LOCL 


19.  The  angle  ACB  is  any  inscribed  an- 
gle in  a  given  segment  of  a  circle  ;  ^C  is 
produced  to  P,  making  CP  equal  to  GB ; 
find  the  locus  of  P. 


20.  The  hypotenuse  of  a  right  triangle  is  given  in  magnitude 
and  position  ;  find  the  locus  of  the  centre  of  the  inscribed  circle. 

21.  The  base  BC  of  a  triangle  ABC  is  given  in  position  and 
magnitude,  and  the  vertical  angle  J.  is  of  a  given  magnitude ; 
find  the  locus  of  the  centre  of  the  inscribed  circle. 


22.  From  a  given  point  O,  any  straight  line  OA 
is  drawn  to  a  given  straight  line  MN^  and  OP  is 
drawn  making  a  given  angle  with  OA^  and  such 
that  OP  is  to  OA  in  a  given  ratio ;  find  the  locus 
of  P. 

With  the  same  construction,  if  OP  is  so  taken 
that  the  product  OP.  OA  is  equal  to  a  given  con- 
stant ;  find  the  locus  of  P. 


23.  From  a  given  point  O,  any  straight 
line  OA  is  drawn  to  a  given  circumfer- 
ence, and  OP  is  drawn  making  a  given 
angle  with  OA,  and  such  that  OP  is  to 
OA  in  a  given  ratio  ;  find  the  locus  of  P. 

With  the  same  construction,  if  OP  is 
so  taken  that  the  product  OP.  OA  is 
equal  to  a  given  constant ;  find  the  locus  of  P. 


24.  One  vertex  of  a  triangle  whose  angles  are  given  is  fixed, 
while  the  second  vertex  moves  on  the  circumference  of  a  given 
circle  ;  what  is  the  locus  of  the  third  vertex  ? 

25.  Through  A,  one  of  the  points  of  intersection  of  two  given 
circles,  any  secant  is  drawn  cutting  the  two  circumferences  in  the 
points  B  and  C;  find  the  locus  of  the  middle  point  of  BG. 


A 


MISCELLANEOUS   EXERCISES.  177 


PROBLEMS. 

26.  Describe  a  circle  through  two  given  points  which  lie  outside 
a  given  line,  the  centre  of  the  circle  to  be  in  that  line.  Show 
when  no  solution  is  possible. 

27.  In  a  given  circle,  inscribe  a  chord  of  a  given  length  which 
produced  shall  be  tangent  to  another  given  circle. 

28.  Through  P,  one  of  the  points  of  intersection  of  two  circum- 
ferences, draw  a  straight  line,  terminated  by  the  circumferences, 
which  shall  be  bisected  in  P. 

29.  Through  one  of  the  points  of  intersection  of  two  circum- 
ferences draw  a  straight  line,  terminated  by  the  circumferences, 
which  shall  have  a  given  length. 

30.  In  a  given  triangle  ABC,  to  inscribe 
a  parallelogram  DEFO,  such  that  the  adja- 
cent sides  DE  and  DO  shall  be  in  a  given 
ratio  and  contain  a  given  angle. 


31.  Construct  a  triangle,  given  its  base,  the  ratio  of  the  other 
two  sides,  and  one  angle. 

32.  To  determine  a  point  in  a  given  arc  of  a  circle,  such  that 
the  chords  drawn  from  it  to  the  extremities  of  the  arc  shall  have 
a  given  ratio. 

33.  To  find  a  point  within  a  given  triangle,  such  that  the  three 
straight  lines  drawn  from  it  to  the  vertices  of  the  triangle  shall 
make  three  equal  angles  with  each  other. 

34.  Inscribe  a  trapezoid  in  a  given  circle,  knowing  its  area  and 
the  common  length  of  its  inclined  sides. 

35.  To  construct  a  triangle,  given  one  angle,  the  side  opposite 
to  that  angle,  and  the  area  (equal  to  that  of  a  given  square). 

36.  Divide  a  given  circle  into  a  given  number  of  equivalent 
parts,  by  concentric  circumferences. 

Also,  divide  it  into  a  given  number  of  parts  proportional  to 
given  lines,  by  concentric  circumferences. 


178  ELEMENTS  OF  GEOMETRY. 

37.  A  circle  being  given,  to  find  a  given  number  of  circles  whose 
radii  shall  be  proportional  to  given  lines,  and  the  sum  of  whose 
areas  shall  be  equal  to  the  area  of  the  given  circle. 

38.  In  a  given  equilateral  triangle,  inscribe  three  equal  circles 
tangent  to  each  other  and  to  the  sides  of  the  triangle. 

Determine  the  radius  of  these  circles  in  terms  of  the  side  of  the 
triangle. 

39.  In  a  given  circle,  inscribe  three  equal  circles  tangent  to  each 
other  and  to  the  given  circle. 

Determine  the  radius  of  these  circles  in  terms  of  the  radius  of 
the  given  circle. 


NUMERICAL  EXAMPLES. 


Note.— The  following  approximate  values  are  close  enough  for 
ordinary  purposes  :  tt  =  2^2^  ^2  =  ||,  |/8  =  |f,  /S  =  ff.  Radius 
of  earth  =  3960  miles. 
\^'  40.  The  vertical  angle  of  an  isosceles  triangle  is  36°,  and  the 
length  of  the  base  is  2  feet ;  find  the  base  angles,  the  length  of 
the  bisector  of  a  base  angle,  and  the  length  of  a  side  of  the  given 
triangle.  Ans.  TZ°,  2  feet,  (1  +  V5)  feet. 

I     41.  One  angle  of  a  triangle  is  60°,  the  including  sides  are  8  feet 
and  8  feet ;  find  the  area  and  the  third  side. 

Ans.  61/3  square  feet,  7  feet. 
.^  '  V    42.  The  three  sides  of  a  triangle  are  9  inches,  10  inches,  and  17 
Lo         inches,  its  area  is  36  square  inches  ;  find  the  area  of  the  inscribed 
^^    circle.  Ans.  47r. 

"^  (/    43.  The  adjacent  sides  of  a  parallelogram  are  12  feet  and  14  feet, 
the  area  is  120  square  feet ;  find  the  long  diagonal. 

Ans.  24  feet. 

44.  The  area  of  a  right  triangle  is  6  square  feet,  the  length  of 
the  hypotenuse  is  5  feet ;  find  the  other  sides. 

Ans.  3  feet,  4  feet. 

45.  ©btain  a  formula  connecting  the  length  of  a  chord  ^,  its 
distance  from  the  centre  c?,  and  the  radius  r. 

Ans.  -  =  r^  —  cP, 
4 

46.  Obtain  a  formula  for  the  length  ^  of  a  common  tangent  to 
two  circles,  given  tlie  radii  r,  r'',  and  the  distance  between  the 
centres  d.  Ans.  {r  —  r^y  4-^  =  ^2  for  external  tangent. 

[r  -f  r^f  +  ^  =  c^  for  internal  tangent. 

47.  Through  what  angle  does  the  hour-hand  of  a  clock  move  in 
1  hour?  in  1  minute?  Through  what  angle  does  the  minute- 
hand  move  in  1  minute  ? 

What  angle  do  the  hands  of  a  clock  make  with  each  other  at 
ten  minutes  past  three  ?  at  quarter  of  six  ?        Ans.  35°,  97°  30^. 

/  179 


180  ELEMENTS  OF  GEOMETRY. 

48.  Two  secants  cut  each  other  without  a  circle,  the  intercepted 
arcs  are  12°  and  48°  ;  what  is  the  angle  between  the  secants  ? 

Two  chords  intersect  within  the  circle,  a  pair  of  opposite  inter- 
cepted arcs  are  12°  and  48°  ;  what  is  the  angle  between  the  chords  ? 

49.  Tw^o  tangents  make  with  each  other  an  angle  of  60° ;  re- 
quired the  lengths  of  the  arcs  into  which  their  points  of  contact 
divide  the  circle,  given  radius  equals  7  inches. 

Ans.  14|  inches,  29^  inches. 

60.  A  swimmer  whose  eye  is  at  the  surface  of  the  water  can 
just  see  the  top  of  a  stake  a  mile  distant ;  the  stake  proves  to  be 
8  inches  out  of  water  ;  required  the  radius  of  the  earth. 

Am.  3960  miles. 

61.  A  passenger  standing  on  the  deck  of  a  steamer  about  to 
start  observes  that  his  eye  is  on  a  level  with  the  top  of  the  wharf, 
which  he  knows  to  be  12  feet  high  ;  when  they  have  steamed  8^ 
miles  the  wharf  disappears  below  the  horizon  ;  required  the  radius 
of  the  earth.  Ans.  3974  miles. 

62.  How  many  miles  is  the  light  of  a  light-house  150  feet  high 
visible  at  sea  ?  A7is.  15. 

63.  On  approaching  Portland  from  the  sea,  Mount  Washington 
is  first  visible  12  miles  from  shore ;  Portland  is  85  miles  from 
Mount  Washington  ;  required  the  height  of  the  mountain. 

Ans.  6270  feet. 

64.  The  latitude  of  Leipsic  is  51°  21^  that  of  Venice  45°  26^  and 
Venice  is  due  south  of  Leipsic ;  how  many  miles  are  they  apart? 
Use  4000  miles  as  the  earth's  radius.  Ans.  413  miles. 

66.  The  latitude  of  the  Peak  of  Teneriffe  is  about  30°  N. ;  the 
rising  sun  shines  on  its  summit  on  the  21st  of  March  9  minutes 
before  it  shines  on  its  base ;  required  the  height  of  the  mountain. 

Ans.  About  12,000  feet. 

66.  A  quarter-mile  running-track  10  feet  wide,  with  straight 
parallel  sides  and  semicircular  ends,  is  to  be  laid  out  in  a  rectan- 
gular field  220  feet  wide.  How  long  must  the  field  be  in  order 
that  a  runner,  keeping  in  the  middle  of  the  track,  may  have  one- 
quarter  of  a  mile  to  cover?  how  much  can  he  gain  by  keeping 
close  to  the  inner  edge  of  the  track  ?  what  is  the  area  of  the  field  ? 
of  the  portion  encircled  by  the  track?  of  the  track  itself? 

Ans.  550  feet ;  31f  feet ;  121,000  square  feet ;  97,428^  square  feet ; 
13,200  square  feet. 

67.  The  fly-wheel  of  an  engine  is  connected  by  a  belt  with  a 
smaller  wheel  driving  the  machinery  of  a  mill.    The  radius  of  the 


NUMERICAL   EXAMPLES. 


181 


fly-wheel  is  7  feet ;  of  the  small  wheel,  21  inches.  How  many 
revolutions  does  the  small  wheel  make  to  one  of  the  fly-wheel  ? 
The  distance  between  the  centres  of  the  two  wheels  is  10^  feet. 
What  is  the  length  of  the  connecting  band  ? 

Ans.  51  feet  2  inches. 

58.  If  from  each  vertex  of  a  regular  polygon  as  a  centre,  with 
a  radius  equal  to  one-half  the  side,  an  arc  is  described  outward 
from  side  to  side  of  the  polygon,  an  ornamental  figure  much 
used  in  architecture  is  formed.  Such  a  figure  formed  on  a  polygon 
of  numerous  sides  is  often  used  as  a  rose-window. 

The  figure  bounded  by  three  arcs  is  called  a  trefoil ;  by  four 
arcs,  a  quatre-foil ;  by  five  arcs,  a  cinque-foil. 


Find  the  area  of  a  tre- 
foil, given  the  distance  be- 
tween the  centres  of  adja- 
cent arcs  equal  to  21  inches. 

Ans.  7.338  square  feet. 


69.  A  rose-window 
of  six  lobes  is  to  be 
placed  in  a  circular 
space  42  feet  in  diam- 
eter. How  many 
square  feet  of  glass 
will  it  contain  ? 

Ans.  1123.8  square 
feet. 


SYLLABUS  OF  PLANE  GEOMETRY. 

POSTULATES,  AXIOMS,  AISTD  THEOKEMS. 


BOOK  I. 
Postulate  I. 

Through  any  two  points  one  straight  line,  and  only  one,  can 

be  drawn. 

Postulate  II. 

Through  a  given  point  one  straight  line,  and  only  one,  can  be 
drawn  having  any  given  direction. 

Axiom  I. 

A  straight  line  is  the  shortest  line  that  can  be  drawn  between 
two  points. 

Axiom  II. 

Parallel  lines  have  the  same  direction. 

Proposition  I. 

At  a  given  point  in  a  straight  line  one  perpendicular  to  the  line 
can  be  drawn,  and  but  one. 

Corollary.  Tiirough  tlie  vertex  of  any  given  angle  one  straight 
line  can  be  drawn  bisecting  the  angle,  and  but  one. 

Proposition  II. 
All  right  angles  are  equal.      '        -^ 

Proposition  III. 

The  two  adjacent  angles  which  one  straight  line  makes  with 
another  are  together  equal  to  two  right  angles. 

Corollary  I.  The  sum  of  all  the  angles  having  a  common  ver- 
tex, and  formed  on  one  side  of  a  straight  line,  is  two  right  angles. 

Corollary  II.  The  sum  of  all  the  angles  that  can  be  formed 
about  a  point  in  a  plane  is  four  right  angles. 

182 


SYLLABUS   OF   PLANE   GEOMETRY.  183 


Proposition  IV. 

If  the  sum  of  two  adjacent  angles  is  two  right  angles,  their 
exterior  sides  are  in  the  same  straight  line. 

Proposition  V. 

If  two  straight  lines  intersect  each  other,  the  opposite  (or  ver- 
tical) angles  are  eq^ual. 

Proposition  VI. 

Two  triangles  are  equal  when  two  sides  and  the  included  angle 
of  the  one  are  respectively  equal  to-  two  sides  and  the  included 
angle  of  the  other. 

Proposition  VII. 

Two  triangles  are  equal  when  a  side  and  the  two  adjacent 
angles  of  the  one  are  respectively  equal  to  a  side  and  the  two 
adjacent  angles  of  the  other. 

Proposition  VIII. 

In  an  isosceles  triangle  the  angles  opposite  the  equal  sides  are 
equal. 

Corollary.  The  straight  line  bisecting  the  vertical  angle  of  an 
isosceles  triangle  bisects  the  base,  and  is  perpendicular  to  the  base. 

Proposition  IX. 

Two  triangles  are  equal  when  the  three  sides  of  the  one  are 
respectively  equal  to  the  three  sides  of  the  other. 

Proposition  X. 

Two  right  triangles  are  equal  when  they  have  the  hypotenuse 
and  a  side  of  the  one  respectively  equal  to  the  hypotenuse  and  a 
side  of  the  other. 

Proposition  XI. 

If  two  angles  of  a  triangle  are  equal,  the  sides  opposite  to  them 
are  equal,  and  the  triangle  is  isosceles. 


Proposition  XII. 

If  two  angles  of  a  triangle  are  unequal,  the  side  opposite  the 
greater  angle  is  greater  than  the  side  opposite  the  less  angle. 


184  ELEMENTS  OF  GEOMETRY. 

Proposition  XIII. 

If  two  sides  of  a  triangle  are  unequal,  the  angle  opposite  the 
greater  side  is  greater  than  the  angle  opposite  the  less  side. 

Proposition  XIV. 

If  two  triangles  have  two  sides  of  the  one  respectively  equal  to 
two  sides  of  the  other,  and  the  included  angles  unequal,  the  tri- 
angle which  has  the  greater  included  angle  has  the  greater  third 
side. 

Proposition  XV. 

If  two  triangles  have  two  sides  of  the  one  respectively  equal  to 
two  sides  of  the  other,  and  the  third  sides  unequal,  the  triangle 
which  has  the  greater  third  side  has  the  greater  included  angle. 

Proposition  XVI. 

From  a  given  point  without  a  straight  line  one  perpendicular 
can  be  drawn  to  the  line,  and  but  one. 

Proposition  XVII. 

The  perpendicular  is  the  shortest  hne  that  can  be  drawn  from 
a  point  to  a  straight  line. 

Proposition  XVIII. 

If  a  perpendicular  is  erected  at  the  middle  of  a  straight  line, 
then  every  point  on  the  perpendicular  is  equally  distant  from  the 
extremities  of  the  line,  and  every  point  not  on  the  perpendicular 
is  unequally  distant  from  the  extremities  of  the  line. 

Proposition  XIX. 

Every  point  in  the  bisector  of  an  angle  is  equally  distant  from 
the  sides  of  the  angle  ;  and  every  point  not  in  the  bisector  is  un- 
equally distant  from  the  sides  of  the  angle  ;  that  is,  the  bisector 
of  an  angle  is  the  locus  of  the  points  within  the  angle  and  equally 
distant  from  its  sides. 

Proposition  XX, 

A  convex  broken  line  is  less  than  any  ether  line  which  envelops 
it  and  has  the  same  extremities. 


SYLLABUS  OF  PLANE  GEOMETRY.         185 

Proposition  XXI. 

If  two  oblique  lines  drawn  from  a  point  to  a  line  meet  the  line 
at  unequal  distances  from  the  foot  of  the  perpendicular,  the  more 
remote  is  the  greater. 

Proposition  XXII. 

Two  straight  lines  perpendicular  to  the  same  straight  line  are 
parallel. 

Proposition  XXIII. 

Through  a  given  point  one  line,  and  only  one,  can  be  .drawn 
parallel  to  a  given  line. 

Proposition  XXIV. 

When  two  straight  lines  are  cut  by  a  third,  if  the  alternate- 
interior  angles  are  equal,  the  two  straight  lines  are  parallel. 

Corollary  I.  When  two  straight  lines  are  cut  by  a  third,  if  a 
pair  of  corresponding  angles  are  equal,  the  lines  are  parallel. 

Corollary  II.  When  two  straight  lines  are  cut  by  a  third,  if  the 
sum  of  two  interior  angles  on  the  same  side  of  the  secant  line  is 
equal  to  two  right  angles,  the  two  lines  are  parallel. 

Proposition  XXV. 

If  two  parallel  lines  are  cut  by  a  third  straight  line,  the  alter- 
nate-interior angles  are  equal. 

Corollary  I.  If  two  parallel  lines  are  cut  by  a  third  straight 
line,  any  two  corresponding  angles  are  equal. 

Corollary  II.  If  two  parallel  lines  are  cut  by  a  third  straight 
line,  the  sum  of  the  two  interior  angles  on  the  same  side  of  the 
secant  line  is  equal  to  two  right  angles. 

Proposition  XXVI. 

The  sum  of  the  three  angles  of  any  triangle  is  equal  to  two 
right  angles. 

Corollary.  If  one  side  of  a  triangle  is  extended,  the  exterior 
angle  is  equal  to  the  sum  of  the  two  interior  opposite  angles. 

Proposition  XXVII. 

The  sum  of  all  the  angles  of  any  convex  polygon  is  equal  to 
twice  as  many  right  angles,  less  four,  as  the  figure  has  sides. 

16* 


186  ELEMENTS  OF  GEOMETRY. 

Proposition  XXVIII. 

Two  parallelograms  are  equal  when  two  adjacent  sides  and  the 
included  angle  of  the  one  are  equal  to  two  adjacent  sides  and  the 
included  angle  of  the  other. 

Corollary.  Two  rectangles  are  equal  when  they  have  equal 
bases  and  equal  altitudes. 

Proposition  XXIX. 

The  opposite  sides  of  a  parallelogram  are  equal,  and  the  oppo- 
site angles  are  equal. 

Proposition  XXX.  ' 

If  two  opposite  sides  of  a  quadrilateral  are  equal  and  parallel, 
the  figure  is  a  parallelogram. 

Proposition  XXXI. 

If  the  opposite  sides  of  a  quadrilateral  are  equal,  the  figure  is  a 
parallelogram. 

Proposition  XXXII. 

The  diagonals  of  a  parallelogram  bisect  each  other. 


BOOK  II. 
PROPOSITIONS. 

Postulate. 

A  circumference  may  be  described  with  any  point  as  centre  and 
any  distance  as  radius. 

Proposition  I. 

Two  circles  are  equal  when  the  radius  of  the  one  is  equal  to  the 
radius  of  the  other. 

Proposition  II. 

Every  diameter  bisects  the  circle  and  its  circumference. 

Proposition  III. 

In  equal  circles,  or  in  the  same  circle,  equal  angles  at  the  centre 
intercept  equal  arcs  on  the  circumference. 

Corollary.  Conversely,  in  the  same  circle,  or  in  equal  circles, 
equal  arcs  subtend  equal  angles  at  the  centre. 


SYLLABUS  OF  PLANE  GEOMETRY.         187 

Proposition  IV. 

In  equal  circles,  or  in  the  same  circle,  equal  arcs  are  subtended 
by  equal  chords. 

Corollary.  Conversely,  in  equal  circles,  or  in  the  same  circle, 
equal  chords  subtend  equal  arcs. 

Proposition  V. 

In  equal  circles,  or  in  the  same  circle,  the  greater  of  two  un- 
equal arcs  is  subtended  by  the  greater  chord,  the  arcs  being  each 
less  than  a  semi-circumference. 

Corollary.  Conversely,  in  equal  circles,  or  in  the  same  circle 
the  greater  of  two  unequal  chords  subtends  the  greater  arc. 

Proposition  VI. 

The  diameter  perpendicular  to  a  chord  bisects  the  chord  and 
the  arcs  subtended  by  it. 

Corollary  I.  The  perpendicular  erected  at  the  middle  point  of 
a  chord  passes  through  the  centre  of  the  circle. 

Corollary  II.  When  two  circumferences  intersect,  the  straight 
line  joining  their  centres  bisects  their  common  chord  at  right 
angles. 

Proposition  VII. 

In  the  same  circle,  or  in  equal  circles,  equal  chords  are  equally 
distant  from  the  centre  ;  and  of  two  unequal  chords  the  less  is  at 
the  greater  distance  from  the  centre. 

Corollary.  Conversely,  in  the  same  circle,  or  in  equal  circles, 
chords  equally  distant  from  the  centre  are  equal ;  and  of  two 
chords  unequally  distant  from  the  centre,  that  is  the  greater 
whose  distance  from  the  centre  is  the  less. 

Proposition  VIII. 
A  straight  line  cannot  intersect  a  circle  in  more  than  two  points. 

Proposition  IX. 

A  straight  line  tangent  to  a  circle  is  perpendicular  to  the  radius 
drawn  to  the  point  of  contact. 

Corollary  I.  A  perpendicular  to  a  tangent  line  drawn  through 
the  point  of  c(mtact  must  pass  through  the  centre  of  the  circle. 

Corollary  II.  If  two  circumferences  are  tangent  to  each  other, 
their  centres  and  their  point  of  contact  lie  in  the  same  straight 
line. 


V^ 


188  ELEMENTS   OF   GEOMETRY. 


Proposition  X. 

When  two  tangents  to  the  same  circle  intersect,  the  distances 
from  their  point  of  intersection  to  their  points  of  contact  are 
equal. 

Proposition  XI. 

Two  parallels  intercept  equal  arcs  on  a  circumference. 

Doctrine  of  Limits. — Theorem. 

If  two  variables  dependent  upon  the  same  variable  are  so  re- 
lated that  they  are  always  equal,  no  matter  what  value  is  given 
to  the  variable  on  which  they  depend,  and  if,  as  the  independent 
variable  is  changed  in  some  specified  way,  each  of  them  ap- 
proaches a  limit,  the  two  Umits  must  be  absolutely  equal. 

Proposition  XII. 

In  the  same  circle,  or  in  equal  circles,  two  angles  at  the  centre 
are  in  the  same  ratio  as  their  intercepted  arcs. 

Proposition  XIII. 

The  numerical  measure  of  an  angle  at  the  centre  of  a  circle  is 
the  same  as  the  numerical  measure  of  its  intercepted  arc,  if  the 
unit  of  angle  is  the  angle  at  the  centre  which  intercepts  the 
adopted  unit  of  arc. 

Proposition  XIV. 

An  inscribed  angle  is  measured  by  one-half  its  intercepted  arc. 
Corollary.  An  angle  inscribed  in  a  semicircle  is  a  right  angle. 

Proposition  XV. 

An  angle  formed  by  a  tangent  and  a  chord  is  measured  by  one- 
half  the  intercepted  arc. 

Proposition  XVI. 

An  angle  formed  by  two  chords  intersecting  within  the  circum- 
ference is  measured  by  one-half  the  sum  of  the  arcs  intercepted 
between  its  sides  and  between  the  sides  of  its  vertical  angle. 

Proposition  XVII. 

An  angle  formed  by  two  secants  intersecting  without  the  cir- 
cumference is  measured  by  one-half  the  difference  of  the  inter- 
cepted arcs. 


SYLLABUS   OF   PLANE   GEOMETRY.  189 

Proposition  XVIII. 

An  angle  formed  by  a  tangent  and  a  secant  is  measured  by  one- 
half  the  difference  of  the  intercepted  arcs. 

Corollary.  An  angle  formed  by  two  tangents  is  measured  by 
one-half  the  difference  of  the  intercepted  arcs. 


BOOK  III. 

THEOREMS. 

Proposition  I. 

A  parallel  to  the  base  of  a  triangle  divides  the  other  two  sides 

proportionally. 

Proposition  II. 

If  a  straight  line  divides  two  sides  of  a  triangle  proportionally, 
it  is  parallel  to  the  third  side. 

Proposition  III. 
Two  triangles  are  similar  when  they  are  mutually  equiangular. 

Proposition  IV. 

Two  triangles  are  similar  when  an  angle  in  the  one  is  equal  to 
an  angle  in  the  other,  and  the  sides  including  these  angles  are 
proportional. 

Proposition  V. 

Two  triangles  are  similar  when  their  homologous  sides  are 
proportional. 

Proposition  VI. 

If  two  polygons  are  composed  of  the  same  number  of  triangles, 
similar  each  to  each  and  similarly  placed,  the  polygons  are 
similar. 

Proposition  VII. 

Two  similar  polygons  may  be  decomposed  into  the  same  num- 
ber of  triangles,  similar  each  to  each  and.  similarly  placed. 

Proposition  VIII. 

The  perimeters  of  two  similar  polygons  are  in  the  same  ratio  as 
any  two  homologous  sides. 


190  ELEMENTS  OF  GEOMETRY. 

Proposition  IX. 

If  a  perpendicular  is  drawn  from  the  vertex  of  the  right  angle 
to  the  hypotenuse  of  a  right  triangle  : 

1st.  The  two  triangles  thus  formed  are  similar  to  each  othei 
and  to  the  whole  triangle  ; 

2d.  The  perpendicular  is  a  mean  proportional  between  the  seg^ 
ments  of  tlie  hypotenuse  ; 

3d.  Each  side  about  the  right  angle  is  a  mean  proportional  be- 
tween the  whole  hypotenuse  and  the  adjacent  segment. 

Corollary.  If  from  any  point  in  the  circumference  of  a  circle  a 
perpendicular  is  let  fall  upon  a  diameter,  the  perpendicular  is  a 
mean  proportional  between  the  segments  of  the  diameter. 

Proposition  X. 

The  square  of  the  length  of  the  hypotenuse  of  a  right  triangle 
is  the  sum  of  the  squares  of  the  lengths  of  the  other  two  sides, 
the  three  lengths  being  expressed  in  terms  of  the  same  unit. 

Proposition  XI. 

If  two  chords  intersect  within  a  circle,  their  segments  are  re- 
ciprocally proportional. 

Proposition  XII. 

If  two  secants  intersect  without  a  circle,  the  whole  secants  and 
their  external  segments  are  reciprocally  proportional. 

Corollary.  If  a  tangent  and  a  secant  intersect,  the  tangent  is  a 
mean  proportional  between  the  whole  secant  and  its  external 
segment. 


BOOK  lY. 

THEOREMS. 

Proposition  I. 

Parallelograms  having  equal  bases  and  equal  altitudes  are 
equivalent. 

Corollary.  Any  parallelogram  is  equivalent  to  a  rectangle 
having  the  same  base  and  the  same  altitude. 


SYLLABUS  OF  PLANE  GEOMETRY.         191 

Proposition  II. 

Two  rectangles  having  equal  altitudes  are  to  each  other  as  their 
bases. 

Corollary.  Two  rectangles  having  equal  bases  are  to  each  other 
as  their  altitudes. 

Proposition  III. 

Any  two  rectangles  are  to  each  other  as  the  products  of  their 
bases  by  their  altitudes. 

Proposition  IV. 

The  area  of  a  rectangle  is  equal  to  the  product  of  its  base  and 
altitude. 

Proposition  V. 

The  area  of  a  parallelogram  is  equal  to  the  product  of  its  base 
and  altitude. 

Proposition  VI. 

The  area  of  a  triangle  is  equal  to  half  the  product  of  its  base 
and  altitude. 

Corollary  I.  A  triangle  is  equivalent  to  one-half  of  any  paral- 
lelogram having  the  same  base  and  the  same  altitude. 

Corollary  II.  Triangles  having  equal  bases  and  equal  altitudes 
are  equivalent. 

Corollary  III.  Triangles  having  equal  altitudes  are  to  each 
other  as  their  bases,  and  triangles  having  equal  bases  are  to  each 
other  as  their  altitudes. 

Proposition  VII. 

The  area  of  a  trapezoid  is  equal  to  the  product  of  its  altitude  by 
half  the  sum  of  its  parallel  bases. 

Proposition  VIII. 

Similar  triangles  are  to  each  other  as  the  squares  of  their  homol- 
ogous sides. 

Proposition  IX. 

Similar  polygons  are  to  each  other  as  the  squares  of  their  homol- 
ogous sides. 

Proposition  X. 

The  square  described  upon  the  hypotenuse  of  a  right  triangle 
is  equivalent  to  the  sum  of  the  squares  described  on  the  other 
two  sides. 


192  ELEMENTS  OF  GEOMETRY. 

BOOK  V. 

THEOREMS. 

Proposition  I. 

If  the  circumference  of  a  circle  be  divided  into  any  number  of 
equal  parts,  the  chords  joining  the  successive  points  of  division 
form  a  regular  polygon  inscribed  in  the  circle  ;  and  the  tangents 
drawn  at  the  points  of  division  form  a  regular  polygon  circum- 
scribed about  the  circle. 

Corollary  I.  If  the  vertices  of  a  regular  inscribed  polygon  are 
joined  with  the  middle  points  of  the  arcs  subtended  by  the  sides 
of  the  polygon,  the  joining  Unes  will  form  a  regular  inscribed 
polygon  of  double  the  number  of  sides. 

Corollary  II.  If  at  the  middle  points  of  the  arcs  joining  ad- 
jacent points  of  contact  of  the  sides  of  a  regular  circumscribed 
polygon  tangents  are  drawn,  a  regular  circumscribed  polygon 
of  double  the  number  of  sides  will  be  formed. 

Proposition  II. 

A  circle  may  be  circumscribed  about  any  regular  polygon,  and 
a  circle  may  also  be  inscribed  in  it. 

Proposition  III. 

Regular  polygons  of  the  same  number  of  sides  are  similar. 

Corollary.  The  perimeters  of  regular  polygons  of  the  same 
number  of  sides  are  to  each  other  as  the  radii  of  the  circumscribed 
circles,  or  as  the  radii  of  the  inscribed  circles ;  and  their  areas  are 
to  each  other  as  the  squares  of  these  radii. 

Proposition  IV. 

The  area  of  a  regular  polygon  is  equal  to  half  the  product  of 
Its  perimeter  and  apothem. 

Proposition  V. 

An  arc  of  a  circle  is  less  than  any  line  which  envelops  it  and 
has  the  same  extremities. 

Corollary,  The  circumference  of  a  circle  is  less  than  the  perim- 
eter of  any  polygon  circumscribed  about  it. 


SYLLABUS  OF  PLANE  GEOMETRY.         193 

Proposition  VI. 

If  the  number  of  sides  of  a  regular  polygon  inscribed  in  a  circle 
be  increased  indefinitely,  the  apothem  of  the  polygon  will  ap- 
proach the  radius  of  the  circle  as  its  limit. 

Proposition  VII. 

The  circumference  of  a  circle  is  the  Umit  which  the  perimeters 
of  regular  inscribed  and  circumscribed  polygons  approach  when 
the  number  of  their  sides  is  increased  indefinitely  ;  and  the  area 
of  the  circle  is  the  limit  of  the  areas  of  these  polygons. 

Proposition  VIII. 

The  circumferences  of  two  circles  are  to  each  other  as  their 
radii,  and  their  areas  are  to  each  other  as  the  squares  of  their 
radii. 

Corollary  I.  The  circumferences  of  circles  are  to  each  other  as 
their  diameters,  and  their  areas  are  to  each  other  as  the  squares 
of  their  diameters. 

Corollary  II.  The  ratio  of  the  circumference  of  a  circle  to  its 
diameter  is  constant. 

Proposition  IX. 

The  area  of  a  circle  is  equal  to  half  the  product  of  its  circum- 
ference by  its  radius. 

Corollary.  The  area  of  a  circle  is  equal  to  the  square  of  its 
radius  multiplied  by  the  constant  number  tt. 


GEOMETRY  OE  SPACE. 


In  Plane  Geometry  we  have  considered  merely  figures  com- 
posed of  lines  and  points,  all  of  which  are  supposed  to  lie  in 
the  same  plane  (v.  Introduction,  5  and  6),  and  in  the  proposi- 
tions and  definitions  of  the  preceding  five  books  it  has  been 
tacitly  assumed  that  the  figures  in  question  are  plane  figures. 
In  many  of  the  propositions  and  definitions  this  limitation  is 
essential  to  the  truth  of  the  proposition ;  for  example.  Propo- 
sitions I.  and  XXII.,  Book  I.,  and  Definition  20,  Book  II.  In 
others  the  demonstration  given  is  inconclusive  without  the 
limitation  in  question,  although  the  proposition  is  true  even 
when  the  limitation  is  removed;  for  example.  Exercise  1, 
Proposition  XXIII.,  Book  I.  While  in  propositions  concern- 
ing equal  polygons,  which  depend  for  their  proof  directly  or 
indirectly  upon  a  superposition  of  one  polygon  upon  the 
other,  the  limitation  is  obviously  of  no  importance;  for  ex- 
ample. Propositions  YI.,  YII.,  and  IX.,  Book  I.  It  is,  then, 
important,  when  we  use  the  theorems  of  Plane  Geometry  in 
proving  theorems  of  the  Geometry  of  Space,  to  satisfy  our- 
selves that  they  are  still  true  in  the  figures  with  which  we  are 
concerned. 


194 


BOOK  YI 

THE   PLANE.     POLYBDRAL   ANGLES. 

1.  Definition.  A  plane  has  already  been  defined  as  a  surface 
such  that  the  straight  line  joining  any  two  points  in  it  lies 
wholly  in  the  surface. 

JIT 

Thus,  the  surface  ikfiV"  is  a  plane,  if,  A  / 

and  B  being  any  two  points  in  it,  the        /—4- ?- 

straight  line  AB  lies  wholly  in  the  sur- 
face. 

The  plane  is  understood  to  be  indefinite  in  extent,  so  that, 
however  far  the  straight  line  is  produced,  all  its  points  lie 
in  the  plane.  But  to  represent  a  plane  in  a  diagram,  we 
are  obliged  to  take  a  limited  portion  of  it,  and  we  usually 
represent  it  by  a  parallelogram  supposed  to  lie  in  the  plane. 

2.  Definition.  A  plane  is  said  to  be  determined  by  given  lines 
or  points  when  one  plane,  and  only  one,  can  be  drawn  con- 
taining the  given  lines  or  points. 

PKOPOSITION  I.— THEOREM. 

K^^Through  any  given  straight  line  a  plane  may  be  passed ; 
hut  the  line  will  not  determine  the  plane. 

Let  AB  be  a  given  straight  line. 
A  straight  line  may  be  drawn  in 
any  plane,  and  the  position  of  that 
plane  may  be  changed  until  the 

line  drawn  in  it  is  brought  into  coincidence  with  AB.     We 
shall  then  have  a  plane  passed  through  AB ;  and  this  plane 

195 


r  ~            J 

7^ 

!.-'' 

'/ 

aJ.                        /I 

B 

L-                       1 

r|o  ^  ELEMENTS   OF   GEOMETRY. 

may  be  turned  upon  AB  as  an  axis,  and  made  to  occupy- 
as  many  different  positions  as  we 
choose,  and  as  in  each  of  these 
positions  it  is  a  plane  through  AB, 


we  may  have  as  many  planes  as  lL 

we    choose    through  AB ;    conse- 
quently AB  does  not  determine  (2)  a  plane. 


PBOPOSITION  II.— THEOKEM. 

/       4.  ^  plane  is  determined^  1st,  by  a  straight  line  and  a  point 
/      without  that  line ;  2d,  by  two  intersecting  straight  lines ;  3d,  by 
7  three  points  not  in  the  same  straight  line ;  4th,  by  two  parallel 
!    straight  lines. 


Ist.  Through  a  given  line  AB  a 
plane  may  be  passed,  and  may  then  /  <?  ^ 

be  turned  upon  AB  as  an  axis,  until 
it  contains  a  given  point  G.  If  it  is 
then  turned  by  the  smallest  amount 

in  either  direction,  it  ceases  to  contain  C.  Therefore  on© 
plane,  and  only  one,  can  be  drawn  containing  a  given  line 
and  a  given  point  without  that  line. 

2d.  Through  one  of  the  lines  ABj  and  a  point  C  of  the 
other,  one  plane,  and  only  one,  can  be 
drawn.    This  plane  will  contain  AG 
(1),  and  no  other  plane  through  AB 
will. 

3d.  If  three  points.   A,   Bj  0,  are 
given,    any    plane    containing    them 

must  contain  one  of  them  and  the  line  joining  the  other  two 
(1)  J  and  one,  and  only  one,  such  plane  can  be  drawn. 


BOOK  VI.  197 

4th.  Two  parallels,  AB,  CD,  lie  in  the  same  plane,  by  Defi- 
nition (I.,  5)  ;  there  is,  then,  one  plane 
containing  them.     There  is  only  one, 
for  through  AB  and  a  point  B  of 

CD  only  one  plane  can  be  passed.  ^       J  ^ 

^fo) Corollary.  The  intersection  of 

two  planes  is  a  straight  line.  For  if  two  points  of  the  in- 
tersection be  joined  by  a  straight  line,  that  line  must  lie  in 
both  planes^  by  (1) ;  and  no  point  outside  of  this  line  can  be 
common  to  the  two  planes,  by  Proposition  II. ;  therefore  the 
straight  line  in  question  is  the  line  of  intersection  of  the  two 
planes.  ^ 

PERPENDICULARS  AND  OBLIQUE  LINES  TO  PLANES. 

6.  Definition.  A  straight  line  is  perpendicular  to  a  plane 
when  it  is  perpendicular  to  every  straight  line  drawn  in  the 
plane  through  its  foot ;  that  is,  through  the  point  in  which  it 
meets  the  plane. 

In  the  same  case,  the  plane  is  said  to  be  perpendicular  to 

the  line. 

PROPOSITION  III.— THEOREM. 

Cy  From  a  given  point  without  a  plane,  one  perpendicular  to 

the  plane  can  he  drawn,  and  hut  one ;  and  the  perpendicular  is 

the  shortest  line  that  can  he  drawn  from  the  point  to  the  plane. 

Let  A  be  the  given  point,  and  JOT  the 
plane.  Consider  the  various  lines  that  can 
be  drawn  from  A  to  MN.  These  lines  are 
obviously  not  all  of  the  same  length  ;  there 
must,  then,  be  among  them  either  one 
minimum  line,  or  a  set  of  equal  shortest 
lines.  There  cannot  be  a  set  of  equal 
shortest  lines.     For,  suppose  that  AB  and  AB'  are  two  such 

17* 


.  y 

^ 

^^ 

p 

198 


ELEMENTS   OF   GEOMETRY. 


-  J 

- 

^^ 

p 

lines.  Join  BB'.  Then,  since  AB  and  AB'' Sire  equal  lines 
drawn  from  A  to  BB\  they  cannot  be  per- 
pendicular to  BB'  (I., 'Proposition  XYI.), 
and  consequently  they  are  longer  than  the 
perpendicular  AC  from  A  to  BB\  by  1., 
Proposition  XYII.,  which  is  contrary  to 
the  hypothesis  that  they  were  shorter 
than  any  other  lines  that  could  be  drawn  ^ 

from  A  to  J/iV.  There  is  therefore  one, 
and  but  one,  minimum  line  from  A  to  the  plane/  Let  AP  be 
that  minimum  line ;  then  AP  is  perpendicular  to  any  straight 
line  BF  drawn  in  the  plane  through  its  foot  P.  For,  in  the 
plane  of  the  lines  AP  and  JEF,  ^P  is  the  shortest  line  that 
can  be  drawn  from  A  to  any  point  in  FF,  since  it  is  the 
shortest  line  that  can  be  drawn  from  J.  to  any  point  in  the 
plane  MN;  therefore  AP  is  perpendicular  to  FF  (I.,  Propo- 
sition XYII.).  Thus  AP  is  perpendicular  to  any^  that  is,  to 
every,  straight  line  drawn  in  the  plane  through  its  foot,  and 
is  therefore  perpendicular  to  the  plane. 

There  can  be  no  other  perpendicular  from  A  to  the  plane 
MN;  for,  if  there  were,  both  lines  would  be  perpe-ndicular 
to  the  line  joining  the  points  where  they  met  the  plane,  and 
we  should  have  two  perpendiculars  from  a  point  to  a  line, 
which  is  contrary  to  I.,  Proposition  XYI. 
vj)  Corollary.  At  a  given  point  in  a  plane j  one  perpendicular 
can  be  erected  to  the  plane,  and  but  one. 

Let  ilOTbe  the  plane  and  P 
the  point. 

Let  Jif'iV'  be  any  other 
plane,  A'  any  point  without  it, 
and  A'P'  the  perpendicular 
from  A'  to  this  plane.     Suppose  the  plane  Al'N^  to  be  applied 


BOOK   VI. 


199 


to  the  plane  MN  with  the  point  P'  upon  P,  and  let  AP  be 
the  position  then  occupied  by  the  perpendicular  A'P'.  We 
then  have  one  perpendicular,  J.P,  to  the  plane  MN^  erected 
at  P.  There  can  be  no  other :  for  let  PB  be  another  perpen- 
dicular at  P.  Then  AP  and  PB  are  both  perpendicular  to 
PC,  the  line  of  intersection  of  MN  with  the  plane  deter- 
mined by  the  two  lines  AP  and  BP^  at  the  same  point,  and 
lie  in  the  same  plane  with^  P(7,  ^and  this  is  contrary  to  I.,     j    j 

Proposition  I.  (^  ^lr^li'  ^x^::..--^rar^) 

9.  Scholium.  By  the  distance  of  a  point  from  a  plane  is 
meant  the  shortest  distance;  hence  it  is  the  perpendicular 
distance  from  the  point  to  the  plane. 


EXERCISES. 

1.  Theorem. — Oblique  lines  drawn  from  a  point  to  a  plane, 
and  meeting  the  plane  at  equal  distances  from  the  foot  of  the 
perpendicular,  are  equal  i  and  of  two  oblique  lines  meeting  the 
plane  at  unequal  distances  from  the  foot  of  the  perpendicular 
the  more  remote  is  the  greater. 


2.  Theorem. — Equal  oblique  lines 
from  a  point  to  a  plane  meet  the 
plane  at  equal  distances  from  the  foot 
of  the  perpendicxdar ;  and  of  two  un- 
equal oblique  lines  the  greater  meets 
the  plane  at  the  greater  distance  from 
the  foot  of  the  perpendicular. 


H 


200  ELEMENTS  OF   GEOMETRY. 


PROPOSITION  IV.— THEOREM. 

(^O)  If  a  straight  line  is  perpendicular  to  each  of  two  straight 
lines  at  their  point  of  intersection^  it  is  perpendicular  to  the  plane 
of  those  lines. 

Let  AP  be  perpendicular  to  PB  and 
P(7,  at  their  intersection  P ;  then  AP  is 
perpendicular  to  the  plane  MN  which 
contains  those  lines. 

For,  let  PD  be   any  other  straight 
line  drawn  through  P  in  the  plane  MN. 
Draw  any  straight  line  BDG  intersect- 
ing PB,  PC,  PD,  in  J5,  (7,  D;  produce  W 
AP  to  A',  making  PA'  =  PA,  and  join                a^ 
A  and  A'  to  each  of  the  points  B,  C,  D, 

Since  BP  is  perpendicular  to  AA',  at  its  middle  point,  we 
have  BA  =  BA'  (I.,  Proposition  XYIII.),  and  for  a  like 
reason  CA  =  GA' ;  therefore  the  triangles  ABC,  A'BC,  are 
equal  (I.,  Proposition  IX.),  and  the  angle  ABD  is  equal  to 
the  angle  A'BD.  The  triangles  ABD  and  A'BD  are  equal 
(I.,  Proposition  YI.),  and  AD  =  A'D.  Hence  the  triangles 
APD  and  A'PD  are  equal  (I.,  Proposition  IX.).  Therefore 
the  adjacent  angles  APD  and  A'PD  are  equal,  and  PD  is 
perpendicular  to  AP.  AP,  then,  is  perpendicular  to  any^ 
that  is,  to  every,  line  passing,  through  its  foot  in  the  plane 
MNp  and  is  consequently  perpendicular  to  the  plane. 
\l|t  Corollary  I.  At  a  given  point  of  g,  straight  line  one 
plane  can  he  drawn  perpendicular  to  the  line,  and  hut  one. 

Let  AP  be  the  line,  and  P  the  point.  Through  AP  pass 
two  planes,  and  in  each  of  these  planes  draw  through  P  a 
line  perpendicular  to  AP.    The  plane  determined  by  these 


BOOK   VI.  201 

two   lines   is   perpendicular   to  AF  at   P,   by  Proposition 

No  other  perpendicular  plane  can 
be  drawn  through  P,  for,  if  it  could, 
a  plane  containing  AF  would  inter- 
sect the  two  perpendicular  planes 
in  lines  which  would  lie  in  the  same 
plane  with  J.P,  and  be  perpendic- 
ular  to   AF   at   the    same   point,     ^  "  *      ^  4^**v   -^"^ 
which   is  contrary  to  I.,  Proposi-(  \  yj^^-^tj^     ok.^^v^"'^' 
tion  I.                                                                                 *  ^-^'H  * 
X    QS^  Corollary  II.   Through  a  given  point  without  a  straight 
line  one  plane  can  he  drawn  perpendicular  to  the  line,  and  but 
one. 

In  the  plane  determined  by  the  point  and  the  line  draw  a 
perpendicular  from  the  point  to  the  line,  and  through  the 
foot  of  this  perpendicular  draw,  in  any  second  plane  passing 
through  the  given  line,  a  second  perpendicular  to  the  line. 
The  plane  of  these  two  perpendiculars  is  obviously  a  plane 
passing  through  the  given  point  and  perpendicular  to  the 
given  line. 

No  second  perpendicular  plane  can  be  drawn  through  thtj 
given  point,  for  the  plane  determined  by  the  line  and  the 
point  would  cut  the  two  perpendicular  planes  in  lines  which 
would  be  two  perpendicular  lines  from  the  given  point  to  the 
given  line,  which  is  contrary  to  I..  Proposition  XYL .         ,    , 

EXERCISES. 

1.  Theorem. — All  the  perpendiculars  that  can  be  drawn  to  a- 
straight  line  at  the  same  point  lie  in  a  plane  perpendicular  to  the 
line  at  the  point. 


202  ELEMENTS   OF   GEOMETRY. 

2.  Theorem. — If  from  the  foot  of  a 
perpendicular  to  a  plane  a  straight  line 
is  drawn  at  right  angles  to  any  line  of 
the  plane,  and  its  intersection  with  that 
line  is  joined  to  any  point  of  the  perpen- 
dicular, this  last  line  will  be  perpendic- 
ular to  the  line  of  the  plane. 

PARALLEL  STRAIGHT  LINES  AND  PLANES. 

13.  Definitions.  A  straight  line  is  parallel  to  a  plane  when  it 
cannot  meet  the  plane,  though  both  be  indefinitely  produced. 

In  the  same  case,  the  plane  is  said  to  be  parallel  to  the  line. 

Two  planes  are  parallel  when  they  do  not  meet,  both  being 
indefinite  in  extent. 

Jy.   PROPOSITION  v.— THEOREM. 

(14)  Two  lines  in  space  having  the  same  direction  are  par- 
allel. 


Let  AB  and  CD  be  two  lines  having     

.  the  same  direction.    Through  AB  and  any 
point  E  of  CD  pass  a  plane,  and  in  this     ^         e  d 

plane  draw  through  E  a  line  parallel  to 
y'^^.y^  '^^'     This  line  will  have  the   same    direction   as  AB  (I., 
'^^^^/^^jSp^iom  II.),  and  consequently  the  same  direction  as  CD,  and 
^>-^^  must  therefore  coincide  with  CD,  by  I.,  Postulate  11.    Hence,-      ^ 
AB  and  CD  are  parallel,    (f  "^^  T''-    ff  •  ^  ^•■'  f '^-^.  tT^l^ 

15.  Corollary.  Two  lines  paraltet  to  the  seme  lirie  are  par- 
allel to  each  other.     For  they  have  the  same  direction. 

^..^^  PROPOSITION  VI.— THEOREM. 

<ISJ  If  two  straight  lines  are  parallel,  every  plane  passed 
through  one  of  them  and  not  coincident  with  the  plane  of  the 
parallels  is  parallel  to  the  other. 


BOOK   VI. 


203 


31 


Let  AB  and  CD  be  parallel  lines,  and  MW  any  plane  passed 
through  CD  ;  then  the  line  AB  and 
the  plane  MN  are  parallel. 

For  the  parallels  AB,  CD,  are  in 
the  same  plane,  A  CDB,  which  in- 
tersects the  plane  MN  in  the  line 
CD;   and   since  AB   cannot  leave 

this  plane,  if  it  meets  MN  at  all  it  must  meet  it  in  some 
point  common  to  the  two  planes;  that^i^,  in  some  point  of 
CD,  which  is  contrary  to  the  hypothesis  that  AB  and  CD 
are  parallel. 

17.  Corollary  I.   Through  any  given  straight  line  a  plane 
can  be  passed  parallel  to  any  other  given  straight  line. 

Let  UK  and  AB  be  the  two  given 
lines.  In  the  plane  determined  by  AB 
and  any  point  S  of  UK  let  HL  be 
drawn  parallel  to  AB ;  then  the  plane 
MNj  determined  by  SK  and  HL,  is  par- 
allel to  AB,  by  Proposition  YI. 

18.  Corollary  II.  Through  any  given  point  a  plane  can  Ir* 
passed  parallel  to  any  two  given  straight  lines  in  space. 

Let  0  be  the  given  point,  and  AB 
and  CD  the  given  straight  lines.  In 
the  plane  determined  by  the  given 
point  0  and  the  line  AB  let  a 06  be 
drawn  through  0  parallel  to  AB ; 
and  in  the  plane  determined  by  the 
point  0  and  the  line  CD  let  cOd  be 
drawn  through  0  parallel  to  CD; 

then  the  plane  determined  by  the  lines  ab  and  cd  is  parallel 
to  each  of  the  lines  AB  and  CD,  by  Proposition  VI. 


--A 


N 


1 


204  ELEMENTS   OF   GEOMETRY. 

EXERCISE. 

Theorem. — If  a  straight  line  and  a  plane  are  parallel^  the 
intersection  of  the  plane  and  a  plane  passed  through  the  given 
line  is  parallel  to  the  given  line.    (v.  Figure  of  Proposition  VI.) 

PROPOSITION  VII.— THEOREM. 

(10}  Planes  perpendicular  to  the  same  straight  line  are  parallel 
to  each  other. 

The  planes  MW,  PQ,  perpendicular  to  the  jn 
same  straight  line  AB,  cannot  meet ;  for,  if  / 
they  met,  we  should  have  through  a  point  L 
of  their  intersection  two  planes  perpendic- 
ular to  the  same  straight  line,  which  is  im-  ^ 
possible    (Proposition    IV.,    Corollary    II.);  / 
therefore  these  planes  are  parallel.  ;^    I         q     \(\ 

PROPOSITION  VIII.— THEOREM. 

^2J  The  intersections  of  two  parallel  planes  with  any  third 
plane  are  parallel. 

Let  MW  and  PQ  be  parallel  planes,        r y\ y 

and  AD  any  plane  intersecting  them  \      /^^  '\             \ 

in  the  lines  AB  and  CD;  then  AB  A          \~ 

and  CD  are  parallel.  p  \ \i) 

For  the  lines  AB  and  CD  cannot  \            \    /       \ 

meet,  since  the  planes  in  which  they  ^ ^ ^ 

are  situated  cannot  meet,  and  they 

are  lines  in  the  same  plane  AD  ;  therefore  they  are  parallel. 

EXERCISE. 

Theorem. — Parallel  lines  intercepted  between  parallel  planes 
are  equal. 


BOOK  VI.  205 

PROPOSITION  IX.— THEOREM. 

21.  A  straight  line  perpendicular  to  one  of  two  parallel  planes 
is  perpendicular  to  the  other. 

Let  MW  and  FQ  be  parallel  planes,  and  let 
the  straight  line  AB  be  perpendicular  to  PQ; 
then  it  will  also  be  perpendicular  to  MN. 

For  through  A  draw  any  straight  line  AG 
in  the  plane  JOT,  pass  a  plane  through  AB 
and  AC,  and  let  BD  be  the  intersection  of 
this  plane  with  PQ.  Then  AC  and  BD  are 
parallel  (Proposition  YIII.) ;  but  AB  is  per- 
pendicular to  BD  (6),  and  consequently  also  to  AC;  there- 
fore AB,  being  perpendicular  to  any  line  AC  which  it  meets 
in  the  plane  JfiV",  is  perpendicular  to  the  plane  JOT. 

22.  Corollary.  Through  any  given  point  one  plane  can  be 
passed  parallel  to  a  given  plane,  and  but  one. 

Suggestion.  Drop  a  perpendicular  line  from  the  point  to  the 

plane,  and  then  pass  a  plane  through  the  point  perpendicular 

to  this  line. 

PROPOSITION  X.— THEOREM. 

23.  If  two  angles,  not  in  the  same  plane,  have  their  sides 
respectively  parallel  and  lying  in  the  same  direction,  they  are 
BQual  and  their  planes  are  parallel. 

Let  BAC,  B'A'C,  be  two  angles  lying 
in  the  planes  MN,  M'N' ;  and  let  AB, 
AC,  be  parallel  respectively  to  A'B', 
A!C',  and  in  the  same  directions. 

1st.  The  angles  BAC  and  B'A!C'  are 
equal.  Take  the  distances  A!B'  and  AB 
equal,  and  A!C'  and  AC  equal,  and  join 
BC  and  B' C .  Draw  now  the  lines  AA', 
BB',  and  Cp'.  The  quadrilaterals  AB'  and  AC  are  parallel- 
is 


206 


ELEMENTS  OF   GEOMETRY. 


ograms,  by  I.,  Proposition  XXX.     Therefore  BB'  and  CC 

are  equal  and  parallel  to  AA'.     They  are 

then  equal   and  parallel  to  each   other 

(Proposition  Y.,  Corollary),  and  BC  is  a 

parallelogram.     Hence  BC  =  B'C\  and 

the  triangles  A5(7and  A'B'C  are  equal, 

by   I.,   Proposition    IX.      Consequently 

the  angles  BAG  dmd.  B'A'C  are  equal. 

2d.  The  planes  MN   and  M'N'  are 
parallel.     For  MN  is  parallel  to  A'B' 
and  A'C\  by  Proposition  YI.,  and  therefore,  since  if  it  met 
M'N'  the  line  of  intersection  would  have  to  cut  one  or  the 
other  of  the  intersecting  lines  A!B'^  A!G\  it  is  parallel  to 
M'W, 


. 

?.             1 

r 

M 

i 

^         A' 

W 

N 


PROPOSITION  XI.— THEOREM. 

Yf  one  of  two  parallel  lines  is  perpendicular  to  a  plane, 
the  other  is  also  perpendicular  to  that  plarie. 

Let  AB^  A'B\  be  parallel  lines,  and 
let  AB  be  perpendicular  to  the  plane 
MN;  then  A'B^  is  also  perpendicular 
to  MN. 

For,  let  A  and  A'  be  the  intersections 
of  these  lines  with  the  plane ;  through 

A'  draw  any  line  A'C  in  the  plane  MN,  and  through  A  draw 
AG  parallel  to  A'G'  and  in  the  same  direction.  The  angles 
BAG,  B'A'G',  are  equal  (Proposition  X.)  ;  but  BAG  m  2i  right 
angle,  since  BA  is  perpendicular  to  the  plane;  hence  B'A'G' 
is  a  right  angle ;  that  is,  B'A'  is  perpendicular  to  any  line 
A'G'  drawn  through  its  foot  in  the  plane  MN,  and  is  conse- 
quently perpendicular  to  the  plane.  q 


vi^.. 


V(" 


»*\-  V    ,'  •-'^• 


\     ■-^V^-.■c.I3t>v^5r-v 


X^ 


<^ 


^ 


BOOK  VI.  207 

25.  Corollary.  Two  straight  lines  perpendicular  to  the  same 
plane  are  parallel  to  each  other, 

EXERCISE. 

Theorem. —  Two  parallel  planes  are  everywhere  equally  distant, 

BIEDRAL  ANGLES.— ANGLE  OF  A  LINE  AND  PLANE, 

ETC. 

26.  Definition.  When  two  planes  meet  and  are  terminated 
by  their  common  intersection,  they  form  a  diedral  angle. 

Thus,  the  planes  AE,  AF^  meeting  in  AB^  and 
terminated  by  AB^  form  a  diedral  angle. 

The  planes  AE^  AF,  are  called  the  faceSy  and 
the  line  AB  the  edge,  of  the  diedral  angle. 

A  diedral  angle  may  be  named  by  four  letters, 
one  in  each  face  and  two  on  its  edge,  the  two 
on  the  edge  being  written  between  the  other  two ;  thus,  the 
angle  in  the  figure  may  be  named  DABC. 

When  there  is  but  one  diedral  angle  formed  at  the  same? 
edge,  it  may  be  nariied  by  two  letters  on  its  edge ,-  thus,  in 
the  preceding  figure,  the  diedral  angle  DABC  may  be  named 
the  diedral  angle  AB. 

27.  Definition.  The  angle  CAD  formed  hj  two  straight  lines 
AC,  AD,  drawn,  one  in  each  face  of  the  diedral  angle,  per- 
pendicular to  its  edge  AB  at  the  same  point,  is  called  the 
plane  angle  of  the  diedral  angle. 

EXERCISES. 

1.  Theorem. — All  plane  angles  of  the  same  diedral  angle  are 
equal,     (v.  Proposition  X.) 

2.  Theorem. — If  a  plane  is  drawn  perpendicidar  to  the  edge 
uj  a  diedral  angle,  its  intersections  with  the  faces  of  the  diedral 
angle  form  the  plane  angle  of  the  diedral  angle. 


208 


ELEMENTS   OF   GEOMETRY. 


Br 


c 


D' 


Ff 


28.  A  diedral  angle  DABG  may  be  conceived  to  be  gener- 
ated by  a  plane,  at  first  coincident  with  a  fixed  plane  AE, 
revolving  upon  the  line  A ^  as  an  axis  until  it  comes  into  the 
position  AF.  In  this  revolution  a  straight  line  CA,  perpen- 
dicular to  AB,  generates  the  plane  angle  CAD. 

29.  Definition.  Two  diedral  angles  are  equal  when  they 
can  be  placed  so  that  their  faces  shall  coincide. 

Thus,  the  diedral  angles  CABD, 
G'A'B'D',  are  equal,  if,  when  the 
edge  A'B'  is  applied  to  the  edge 
AB  and  the  face  A'F'  to  the  face 
AF^  the  face  A'E'  also  coincides 
with  the  face  AE. 

Since  the  faces  continue  to  co- 
incide when  produced   indefinitely,  it  is  apparent  that  the 
magnitude  of  the  diedral  angle  does  not  depend  upon  the 
extent  of  its  faces,  but  only  upon  their  relative  position. 

30.  Definition.  Two  diedral  angles  GABD,  DABEj  which 
have  a  common  edge  AB  and  a  common  plane 

BD  between  them,  are  called  adjacent. 

Two  diedral  angles  are  added  together  by 
placing  them  adjacent  to  each  other.  Thus,  the 
diedral  angle  GABE  is  the  sum  of  the  two  diedral 
angles  GABD  and  DABE. 


E' 


,,^ 


31.  Definition.  Two  planes  are  perpen- 
dicular when  the  plane  angle  of  the  di- 
edral angle  which  they  form  is  a  right 
angle.  The  diedral  angle  is  then  called 
a  right  diedral  angle. 


M 


/ 


cA 


VI 


N 


BOOK  VI. 


209 


1/ 


B' 


C 


F' 


PROPOSITION  XII.— THEOREM. 

(^  Two  diedral  angles  are  equal  if  their  plane  angles  are 
equal. 

Let  the  plane  angles  CAD  and 
C'A'D'  of  the  diedral  angles  CABD, 
C'A'B'D',  be  equal;  then  are  the 
diedral  angles  equal. 

For,  superpose  C'A'B'iy  upon 
CABD,   making    the    plane    angle 

C'A'D'  coincide  with  its  equal  CAD ;  then  the  planes  of 
these  angles  will  coincide  (Proposition  II.).  A'B'  and  AB^ 
being  now  perpendicular  to  the  same  plane  at  the  same  point, 
must  coincide  (Proposition  III.,  Corollary) ;  and,  finally,  the 
planes  B'C  and  50  will  coincide,  and  B'D'  and  BD  (Propo- 
sition 11.) .     Therefore  the  diedral  angles  are  equal. 


PROPOSITION  XIII.— THEOREM. 

331  Two  diedral  angles  are  in  the  same  ratio  as.  their 
angles. 

Let  CABD  and  GEFH  be  two         ' 
diedral  angles;  and  let  CAD  and 
GEH  be  their  plane  angles. 

Suppose  the  plane  angles  have 
a  common  measure,  contained  m 
times  in  CAD  and  n  times  in 
GEH;  we  have,  then, 

CAD  ^m 
GEH      n 


plane 


H 


V- ~ 

1   1  II 
1   III 
1   ill 
1   1  1 1 
1   1  II 

F 

a 

-  -1 

V 

^^^^ 

-ij 

Apply  this  measure  to  CAD  and  GEH^  and  through  the 
lines  of  division  and  the  edges  of  the  given  diedral  angles 

o  18* 


210 


ELEMENTS   OF  GEOMETRY. 


pass  planes,  thus  dividing  CABD  into  m  and  GEFII  into  n 
smaller  diedral  angles.  Each  of 
these  small  diedral  angles  has  one 
of  the  parts  into  which  CAD  is 
divided,  or  one  of  the  parts  into 
which  GEH  is  divided,  as  its 
plane  angle,  because  AB  is  per- 
pendicular to  the  plane  of  CAD, 
and  EF  to  the  plane  of  GEH,  by 

Proposition  lY.     These   small  diedral  angles  are,  then,  all 
equal,  by  Proposition  XII.,  and  we  have 


?Fr::rl 


0 

" 

^^-\ 

■\i 

Therefore 


CABD  ^m 
GEFH      n' 

CABD  ^  CAD 
GEFH       GEH' 


The  proof  is  extended  to  the  case  where  the  given  planej 
angles  are  incommensurable,  by  the  method  exemplified  in 
the  proof  of  II.,  Proposition  XII.,  of  III.,  Proposition  I.,  and 
of  IV.,  Proposition  II. 

34.  Scholium.  Since  the  diedral  angle  is  proportional  to  its 
plane  angle  (that  is,  varies  proportionally  with  it),  the  plane 
angle  is  taken  as  the  measure  of  the  diedral  angle,  just  as  an 
arc  is  taken  as  the  measure  of  a  plane  angle.  Thus,  a  diedral 
angle  will  be  expressed  by  45°  if  its  plane  angle  is  expressed 
by  45°,  etc. 


PROPOSITION  XIV.— THEOREM. 

V^Sa  If  a  straight  line  is  perpendicular  to  a  plane,  every  plane 
passed  through  the  line  is  also  perpendicular  to  that  plane. 


BOOK 


VV 


211 


Let  AB  be  perpendicular  to  the  plane  MJSf;  then  any  plane 
PQ,  passed  through  AB,  is  also  perpendic- 
ular to  JfiV. 

For,  at  B  draw  BC^  in  the  plane  MWj 
perpendicular  to  the  intersection  BQ.  Since 
AB  is  perpendicular  to  the  plane  3IW,  it  is 
perpendicular  to  BQ  and  BG;  therefore 
the  angle  ABC  is  the  plane  angle  of  the 
diedral  angle  formed  by  the  planes  PQ  and  MN;  and  since 
the  angle  ABC  is  a  right  angle,  the  planes  are  perpendicular 
to  each  other. 


PROPOSITION  XV.-THEOBEM. 

(^g)  If  two  planes  are  perpendicular  to  each  other,  a  straight 
line  drawn  in  one  of  them,  perpendicular  to  their  intersection,  is 
perpendicular  to  the  other. 

Let  the  planes  PQ  and  JfZV  be  perpendic- 
ular to  each  other ;  and  at  any  point  B  of 
their  intersection  BQ  let  BA  be  drawn,  in 
the  plane  PQ,  perpendicular  to  BQ;  then 
BA  is  perpendicular  to  the  plane  3fN. 

For,  drawing  BC,  in  the  plane  MW,  per- 
pendicular to  BQ,  the  angle  ABC  is  a  right  angle,  since  it  is 
the  plane  angle?  of  the  right  diedral  angle  formed  by  the  two 
planes ;  therefore  AB,  perpendicular  to  the  two  straight  lines 
BQ,  BC,  is  perpendicular  to  their  plane  MW  (Proposition 
IV.). 

37.  Corollary  I.  If  two  planes  are  perpendicular  to  each 
other,  a  straight  line  drawn  through  any  point  of  their  intersec- 
tion perpendicular  to  one  of  the  planes  will  lie  in  the  other,  (v. 
Proposition  III.,  Corollary.) 


212 


ELEMENTS   OF   GEOMETRY. 


38.  Corollary  II.  If  two  planes  are  perpendicular^  a  straight 
line  let  fall  from  any  point  of  one  plane  perpendicular  to  the 
other  will  lie  in  the  first  plane,    (v.  Proposition  III.) 

PROPOSITION  XVI.— THEOREM. 

39.  If  two  intersecting  planes  are  each  perpendicular  to  a 
third  plane,  their  intersection  is  also  perpendicular  to  that  plane. 

Let  the  planes  PQ,  BS,  intersecting  in 
the  line  AB,  be  perpendicular  to  the  plane 
MN ;  then  AB  is  perpendicular  to  the 
plane  M]Sf. 

For,  if  from  any  point  A  of  A  B  a  per- 
pendicular be  drawn  to  MNj  this  perpen- 
dicular will  lie  in  each  of  the  planes  PQ 
and  BS  (Proposition  XY.,  Corollary  II.), 
and  must  therefore  be  their  intersection  AB. 


EE3 


PROPOSITION  XVII.— THEOREM. 

fwi  Through  any  given  straight  line  a  plane  can  be  passed 
perpendicular  to  any  given  plane. 

Let  AB  be  the  given  straight  line,  and 
MW  the  given  plane.  From  any  point  A 
of  AB  let  AC  he  drawn  perpendicular  to 
MN,  and  through  AB  and  A  C  pass  a  plane 
AD.  This  plane  is  perpendicular  to  MN 
(Proposition  XIY.). 

Moreover,  since,  by  Proposition  XY.,  Corollary  IL,  any 
plane  passed  through  AB  perpendicular  to  MN  must  contain 
the  perpendicular  A  C,  the  plane  AD  is  the  only  plane  per- 
pendicular to  MN  that  can  be  passed  through  AB,  unless  AB 
is  itself  perpendicular  to  MN,  in  which  case  every  plai^ 
through  AB  is  perpendicular  to  MN.  ^ 


a 


BOOK    VI. 


213 


EXERCISE. 


Theorem. — The  locus  of  the 
points  equally  distant  from 
two  given  -planes  is  ttie  plane 
bisecting  the  diedral  angle  be- 
tween the  given  planes,  (v. 
I.,  Proposition  XIX.) 


41.  Definitions.  The  projection  of  a 
point  A  upon  a  plane  MN  is  the  foot  a 
of  the  perpendicular  let  fall  from  A 
upon  the  plane. 

The  projection  of  a  line  ABODE . . . 
upon  a  plane  MN  is  the  line  abcde . . . 
containing  the  projections  of  all  the 
points  of  the  line  ABODE  . . .  upon  the 
plane. 

PROPOSITION  XVIII.— THEOREM. 

(42^  The  projection  of  a  straight  line  upon  a  plane  is  a  straight 
line. 

Let  AB  be  the  given  straight  line, 
and  MI^  the  given  plane.  The  plane 
Ab,  passed  through  AB  perpendicular 
to'  the  plane  MN,  contains  all  the  per- 
pendiculars let  fall  from  points  of  AB 

upon  MN  (Proposition  XY.,  Corollary  /V  ^  i^u\t  i    *^ 

II.);  therefore  these  perpendiculars  all  t^t^^  .  ^^^'^a-y.^AJ 

meet  the  plane  MN  in  the  intersection  ab  of  the  perpeiidic-  i   ,  V^ 
ular  plane  with  MN.    The  projection  of  AB  upon  the  plane 
MN  is,  consequently,  the  straight  line  ab. 


214 


ELEMENTS   OF   GEOMETRY. 


43.  Scholium.  The  plane  Ah  is  called  the  projecting  plane  of 
the  straight  line  AB  upon  the  plane  MN. 

PKOPOSITION  XIX.— THEOREM. 

44.  The  acute  angle  which  a  straight  line  makes  with  its  own 
projection  upon  a  plane  is  the  least  angle  which  it  makes  with 
any  line  of  that  plane. 

Let  Ba  be  the  projection  of  the  straight 
line  BA  upon  the  plane  MN,  the  point  B 
being  the  point  of  intersection  of  the  line 
BA  with  the  plane ;  let  BC  be  any  other 
straight  line  drawn  through  B  in  the 
plane;  then  the  angle  ABa  is  less  than 
the  angle  ABC. 

For,  take  BG  =  Ba,  and  join  AC.  In  the  triangles  ABa, 
ABC,  we  have  AB  common,  and  Ba  =  BC;  but  Aa  <i  AC, 
since  the  perpendicular  is  less  than  any  oblique  line ;  therefore 
the  angle  ABa  is  less  than  the  angle  ABC(1.,  Proposition  XY.). 

45.  Definition.  The  acute  angle  which  a  straight  line  makes 
with  its  own  projection  upon  a  plane  is  called  the  inclination 
of  the  line  to  the  plane,  or  the  angle  of  the  line  and  plane. 

46.  Definition.  Two  straight  lines  AB, 
CD,  not  in  the  same  plane,  are  regarded 
as  making  an  angle  with  each  other  which 
is  equal  to  the  angle  between  two  straight 
lines  Ob,  Od,  drawn  through  any  point  O 
in  space,  parallel  respectively  to  the  two 
lines  and  in  the  same  directions. 

47.  From  the  preceding  definition,  it  follows  that  when  a 
straight  line  is  perpendicular  to  a  plane,  it  is  perpendicular  to  all 
the  lines  of  the  plane,  whether  the  lines  pass  through  its  foot 
or  not. 


BOOK  VI.  215 


POLYEDRAL  ANGLES. 


48.  Definition.  When  three  or  more  planes  meet  in  a  com- 
mon point,  they  form  a  polyedral  angle. 

Thus  the  figure  S-ABCD,  formed  by  the 
planes  ASB,  BSC,  CSD,  DSA,  meeting  in 
the  common  point  S^  is  a  polyedral  angle. 

The  point  S  is  the  vertex  of  the  angle ; 
the   intersections   of  the  planes  SA^  SB, 
etc.,  are   its   edges;    the   portions   of   the 
planes  included  between  the  edges  are  its  faces ;  the  angles 
ASB^  BSG,  etc.,  formed  by  the  edges,  are  its  face  angles. 

A  triedral  angle  is  a  polyedral  angle  having  but  three  faces, 
which  is  the  least  number  of  faces  that  can  form  a  polyedral 
angle. 

49.  In  a  polyedral  angle  every  pair  of  adjacent  edges  form 
a  face  angle,  and  every  pair  of  adjacent  faces  form  a  diedral 
angle.  These  face  angles  and  diedral  angles  are  the  parts  of 
the  polyedral  angle. 

50.  Definition.  Two  polyedral  angles  are  equal  when  their 
faces  and  edges  can  be  made  to  coincide,  if  one  angle  is  suit- 
ably superposed  upon  the  other. 

Of  course  it  follows  that  corresponding  parts  of  two  equal 
polyedral  angles  are  equal. 

51.  Definition.  Two  polyedral  angles  are  symmetrical  if  the 
parts  of  one  are  respectively  equal  to  the  parts  of  the  other ; 
but  the  corresponding  parts  succeed  each  other  in  the  two 
angles  in  inverse  order.  When  two  polyedral  angles  are 
(symmetrical,  it  is  impossible  to  superpose  one  upon  the  other 
in  such  a  way  as  to  bring  corresponding  parts  together.  One 
figure  is,  so  to  speak,  right-handed  and  the  other  left-handed. 


216  ELEMENTS  OF  GEOMETRY. 

52.  Definition.  A  polyedral  angle  S-ABCD  is  convex,  when 
any  section,  ABCD,  made  by  a  plane  cutting  all  its  faces,  is 
a  convex  polygon  (I.,  54). 


PROPOSITION  XX.— THEOREM. 

)-The  sum  of  any  two  face  angles  of  a  triedral  angle  is 
greater  than  the  third. 

The  theorem  requires  proof  only  when  the  third  angle 
considered  is  greater  than  each  of  the  others. 

Let  S-ABC  be  a  triedral  angle  in  which 
the  face  angle  ASC  is  greater  than  either 
ASB  or  BSC;  then  ASB  +  BSG  >  ASC. 

For  in  the  face  ASC  draw  SD  making  the 
angle  ASD  equal  to  ASB,  and  through  any 
point  D  of  SB  draw  any  straight  line  ADC 
cutting  SA  and  SG;  take  SB  =  SD,  and  join  AByBG. 

The  triangles  ASD  and  ASB  are  equal,  by  the  construction 
(I.,  Proposition  YI.),  whence  AD  =  AB.  Now,  in  the  tri- 
angle ABC,  we  have 

AB  +  BC>AG, 

and,  subtracting  the  equals  AB  and  AD, 

BC  >  DC; 

therefore,  in  the  triangles  BSC  and  DSC,  we  have  the  angle 
BSC  >  DSC  (I.,  Proposition  XY.),  and  adding  the  equal 
angles  ASB  and  ASD,  we  have  ASB  +  BSC  >-ASC. 


BOOK   VT.  217 

PROPOSITION  XXI.— THEOREM. 

[5^  The  sum  of  the  face  angles  of  any  convex  polyedral  angle 
is  less  than  four  right  angles. 

Let  the  polyedral  angle  S  be  cut  by  a  plane,  making  the 
section  ABCDE^  by  hypothesis,  a  convex 
polygon.     From  any  point  0  within  this 
polygon  draw  OA,  OB,  OC,  OB,  OK 

The  sum  of  the  angles  of  the  triangles 
ASB,  BSC,  etc.,  which  have  the  common 
vertex  S,  is  equal  to  the  sum  of  the  an- 
gles of  the  same  number  of  triangles 
A  OB,  BOO,  etc.,  which  have  the  common 
vertex  0.  But  in  the  triedral  angles  formed  at  A,  B,  0,  etc., 
by  the  faces  of  the  polyedral  angle  and  the  plane  of  the 
polygon,  we  have  (Proposition  XX.) 

SAB  +  SAB  >  EAB, 
SBA  +  SBC>  ABC,  etc.  j 

hence,  taking  the  sum  of  all  these  inequalities,  it  follows  that 
the  sum  of  the  angles  at  the  bases  of  the  triangles  whose 
vertex  is  S  is  greater  than  the  sum  of  the  angles  at  the 
bases  of  the  triangles  whose  vertex  is  0;  therefore  the  sum 
of  the  angles  at  >S^  is  less  than  the  sum  of  the  angles  at  0; 
that  is,  less  than  four  right  angles. 


r> 


218 


ELEMENTS  OF   GEOMETRY. 


*>C? 


PROPOSITION  XXII.-THEOREM. 

55.  If  two  triedral  angles  have  the  three  face  angles  of  the  one 
respectively  equal  to  the  three  face  angles  of  the  other^  the  cor- 
responding diedral  angles  are  equal. 

In  the  triedral  angles  S  and  s,  let  ASB  =  ash,  ASC  =  asc, 
and  BSC  =  bsc ;  then  the  diedral  angle  SA  is  equal  to  the 
diedral  angle  sa. 


\ 


.^  On  the  edges  of  these  angles  take  the  six  equal  distances 

SA,  SB,  SC,  sa,  sb,  sc,  and  draw  AB,  BC,  AC,  ah,  he,  ac.  The 
isosceles  triangles  SAB  and  sab  are  equal,  having  an  equal 
angle  included  by  equal  sides,  hence  AB  =  ah ;  and  for  the 
same  reason,  BG  =  hc,  AC  =  ac ;  therefore  the  triangles 
ABC  and  abc  are  equal. 

At  any  point  D  in  SA,  draw  DE  in  the  face  ASB  and  DF 
in  the  face  ASC,  perpendicular  to  SA;  these  lines  meet  AB 
and  AC,  respectively,  for,  the  triangles  ASB  and  ASC  being 
isosceles,  the  angles  SAB  and  SAC  are  acute;  let  E  and  F 
be  the  points  of  meeting,  and  join  EF.  Now  on  sa  take  sd 
=  SB,  and  repeat  the  same  construction  in  the  triedral 
angle  s. 

The  triangles  ADE  and  ade  are  equal,  since  AB  =  ad,  and 
the  angles  at  A  and  B  are  equal  to  the  angles  at  a  and  d; 
hence  AE  =  ae  and  BE  =  de.    In  the  same  manner  we  have 


BOOK   VI.  219 

AF  =  of  and  DF  =  df.  Therefore  the  triangles  AEF  and 
aef  are  equal  (I.,  Proposition  YI.)j  and  we  have  FF  =  ef. 
Finally,  the  triangles  FDF  and  edf,  being  mutually  equilat- 
eral,  are  equal;  therefore  the  angle  FDF,  which  measures 
the  diedral  angle  SA,  is  equal  to  the  angle  edf,  which  meas- 
ures the  diedral  angle  sa,  and  the  diedral  angles  SA  and  sa 
are  equal  (Proposition  XII.).  In  the  same  manner  it  may 
be  proved  that  the  diedral  angles  SB  and  SO  are  equal  to 
the  diedral  angles  sb  and  sc,  respectively. 

Scholium.  It  follows  that  the  polyedral  angles  >S^  and  s  are 
either  equal  or  symmetrical.  Both  cases  are  represented  in 
the  figure. 


EXERCISES  ON  BOOK  YI. 


THEOREMS. 


1.  If  a  straight  line  AB  is  parallel  to  a  plane  JfJV,  any  plane 
perpendicular  to  the  line  AB  is  perpendicular  to  the  plane  3lJS\ 
(v.  Proposition  VI.,  Exercise.) 

2.  If  a  plane  is  passed  through  one  of  the  diagonals  of  a  paral- 
lelogram, the  perpendiculars  to  this  plane  from  the  extremities 
of  the  other  diagonal  are  equal. 

3.  If  the  intersections' of  a  number  of  planes  are  parallel,  all  the 
perpendiculars  to  these  planes,  drawn  from  a  common  point  in 
space,  lie  in  one  plane. 

Suggestion.  Through  the  common  point  pass  a  plane  perpen- 
dicular to  one  of  the  intersections,  {v.  Proposition  XV.,  Corol- 
lary II.) 

4.  If  the  projections  of  a  number  of  points  on  a  plane  are  in  a 
straight  line,  these  points  are  in  one  plane. 

5.  If  each  of  the  projections  of  a  line  AB  upon  two  intersecting 
planes  is  a  straight  line,  the  line  AB  is  a  straight  line. 

6.  Two  straight  lines  not  in  the  same  plane  being  given  :  1st,  a 
common  perpendicular  to  the  two  lines  can  be  drawn ;  2d,  the 
common  perpendicular  is  the  short-    . 

est  distance  between  the  two  lines.  o         e  d 

Suggestion.  Let  AB  and  CD  be 
the  two  given  lines.  Pass  through 
AB  a  plane  MN  parallel  to  CX>,  and 
through  AB  and  CD  pass  planes 
perpendicular  to  MN.  Their  inter- 
section Co  is  the  required  common 
perpendicular.  CD  and  cd  are  par- 
allel, by  18,  Exercise.  N 

2d.  Any  other  line  BF  joining 
AB  and  CD  is  greater  than  JEIT,  the  perpendicular  from  B  to  cd 
(Proposition  XV.),  and  therefore  greater  than  Cc. 
220 


M 

,1        1 

l'\ 

/i    i    / 

i 

V  " 

'  J 

\ 

7 

BOOK  VI. 


7.  If  two  straight  lines  are  intersected  by 
three  parallel  planes,  their  corresponding 
segments  are  proportional,  {v.  Proposition 
VIII.) 


8.  A  plane  passed  through  the  middle  point  of  the  common 
perpendicular  to  two  straight  lines  in  space,  and  parallel  to  both 
these  lines,  bisects  every  straight  line  joining  a  point  of  one  of 
these  Hues  to  a  point  of  the  other,     {v.  Exercise  7.) 

9.  In  any  triedral  angle,  the  three  planes  bisecting  the  three 
diedral  angles  intersect  in  the  same  straight  line.  {v.  40,  Exer- 
cise.) 

10.  In  any  triedral  angle,  the  three  planes 
passed  through  the  edges  and  the  bisectors 
of  the  opposite  face  angles  respectively  in- 
tersect in  the  same  straight  line. 

Suggestion.  Lay  off  equal  distances  SAj 
SB,  SC,  on  the  three  edges,  and  pass  a  plane 
through  Aj  B,  C.  The  intersections  of  the 
three  planes  in  question  with  ABCsive  the 

medial  lines  of  ABC,  and  have  a  common  intersection,  and  the  line 
joining  this  common  intersection  with  S  lies  in  the  three  planes. 

11.  In  any  triedral  angle,  the  three  planes  passed  through  the 
bisectors  of  the  face  angles,  and  perpendicular  to  these  faces 
respectively,  intersect  in  the  same  straight  line. 

Suggestion.  Use  the  same  construction  as  in  Exercise  10.  Then 
the  intersections  of  the  three  planes  with  ABC  are  perpendicular 
to  the  sides  of  ABC  at  their  middle  points,  and  have  a  common 
intersection. 

12.  In  any  triedral  angle,  the  three  planes 
passed  through  the  edges,  perpendicular  to 
the  opposite  faces  respectively,  intersect  in 
the  same  straight  line. 

Suggestion.  At  any  point  A  of  one  of  the 
edges,  draw  a  plane  ABC  perpendicular  to 
the  edge  SA.  The  intersections  of  the  three 
planes  with  ABC  are  the  perpendiculars 
from  the  vertices  of  ABC^  upon  the  oppo- 
site sides,  and  have  a  common  intersection,   {v.  Proposition  XVI.) 

19* 


222  ELEMENTS   OF   GEOMETRY. 


LOCI. 

13.  Find  the  locus  of  the  points  in  space  which  are  equally 
distant  from  two  given  points. 

14.  Locus  of  the  points  which  are  equally  distant  from  two 
given  straight  lines  in  the  same  plane. 

15.  Locus  of  the  points  which  are  equally  distant  from  three 
given  points. 

16.  Locus  of  the  points  which  are  equally  distant  from  three 
given  planes,    {v.  40,  Exercise.) 

17.  Locus  of  the  points  which  are  equally  distant  from  three 
given  straight  lines  in  the  same  plane. 

\  18.  Locus  of  the  points  which  are  equally  distant  from  the  three 
edges  of  a  tried ral  angle  (Exercise  11). 

19.  Locus  of  the  points  in  a  given  plane  which  are  equally  dis- 
tant from  two  given  points  out  of  the  plane. 

20.  Locus  of  the  points  which  are  equally  distant  from  two 
given  planes,  and  at  the  same  time  equally  distant  from  two 
given  points. 


PROBLEMS. 


In  the  solution  of  problems  in  space,  we  assume, — 1st,  that  a 
plane  can  be  drawn  passing  through  three  given  points  (or  two  in- 
tersecting straight  lines)  and  its  intersections  with  given  straight 
lines  or  planes  determined;  and,  2d,  that  a  perpendicular  to  a 
given  plane  can  be  drawn  at  a  given  point  in  the  plane,  or  from 
a  given  point  without  it.  The  actual  graphic  construction  of  the 
solutions  belongs  to  Descriptive  Geometry. 

21.  Through  a  given  straight  line,  to  pass  a  plane  perpendicular 
to  a  given  plane,    (v.  Proposition  XVII.) 

22.  Through  a  given  point,  to  pass  a  plane  perpendicular  to  a 
given  straight  line. 

Suggestion.  If  the  given  point  is  in  the  given  line,  pass  two 
planes  through  the  given  line,  and  draw  in  each  of  them,  through 
the  given  point,  a  line  perpendicular  to  the  given  line.  The  plane 
determined  by  these  lines  is  the  perpendicular  plane  required. 
{v.  Proposition  IV.) 

If  the  given  point  is  not  in  the  given  line,  pass  a  plane  through 
it  and  the  given  line,  and  in  this  plane,  through  the  given  point, 
draw  a  line  parallel  to  the  given  line.  A  plane  through  the  given 
point,  perpendicular  to  this  second  line,  is  the  plane  required,  (v. 
Proposition  XI.) 


BOOK  VI.  223 

23.  Through  a  given  point,  to  pass  a  plane  parallel  to  a  given 
plane,     (v.  Proposition  IX.,  Corollary.) 

24.  To  determine  that  point  in  a  given  straight  line  which  is 
equidistant  from  two  given  points  not  in  the  same  plane  with 
the  given  line.     (v.  Exercise  13.) 

25.  To  find  a  point  in  a  plane  which  shall  be  equidistant  from 
three  given  points  in  space. 

26.  Through  a  given  point  in  space,  to  draw  a  straight  line 
which  shall  cut  two  given  straight  lines  not  in  the  same  plane. 

Suggestion.  Pass  a  plane  through  the  given  point  and  through 
one  of  the  given  lines  ;  the  line  through  the  given  point  and  the 
point  where  the  plane  cuts  the  second  given  line  is  the  solution 
required. 

27.  Through  a  given  point,  to  draw  a  straight  line  which  shall 
meet  a  given  straight  line  and  the  circumference  of  a  given  circle 
not  in  the  same  plane.     (Two  solutions  in  general.) 

28.  In  a  given  plane  and  through  a  given  point  of  the  plane,  to 
draw  a  straight  line  which  shall  be  perpendicular  to  a  given  line 
in  space. 

Suggestion.  Draw  a  plane  through  the  given  point  and  perpen- 
dicular to  the  given  line.  Its  intersection  with  the  given  plane  is 
the  solution  required. 

29.  Through  a  given  point  ^  in  a  plane,  to  draw  a  straight  line 
AT  in  that  plane,  which  shall  be  at  a  given  distance  PT  from  a 
given  point  P  without  the  plane. 

Suggestion.  Drop  a  perpendicular  from  P  to  the  plane,  and  with 
the  foot  of  this  perpendicular  as  a  centre,  and  with  a  radius  equal 
to  a  side  of  a  right  triangle  whose  hypotenuse  is  PT,  and  whose 
other  side  is  the  length  of  the  perpendicular,  describe  a  circum- 
ference in  the  plane.  A  tangent  from  A  to  this  circumference  is 
the  solution  required,    {v.  12,  Exercise  2.) 

30.  Through  a  given  point  A,  to  draw  to  a  given  plane  M  a 
straight  line  which  shall  be  parallel  to  a  given  plane  N  and  of  a 
given  length. 


BOOK  YIL 

POLYBDRONS. 

1.  Definition.  A  polyedron  is  a  geometrical  solid  bounded 
by  planes. 

The  bounding  planes,  by  their  mutual  intersections,  limit 
each  other,  and  determine  the  faces  (which  are  polygons), 
the  edgeSj  and  the  vertices  of  the  polyedron.  A  diagonal  of  a 
polyedron  is  a  straight  line  joining^  any  two  of  its  vertices 
not  in  the  same  face. 

The  least  number  of  planes  that  can  form  a  polyedral 
angle  is  three ;  but  the  space  within  the  angle  is  indefinite  in 
extent,  and  it  requires  a  fourth  plane  to  enclose  a  finite  por- 
tion of  space,  or  to  form  a  solid ;  hence  the  least  number  of 
planes  that  can  form  a  polyedron  is  four. 

2.  Definition.  A  polyedron  of  four  faces  is  called  a  tetra- 
edron ;  one  of  six  faces,  a  hexaedron ;  one  of  eight  faces,  an 
octaedron ;  one  of  twelve  faces,  a  dodecaedron ;  one  of  twenty 
faces,  an  icosaedron. 

3.  Definition.  A  polyedron  is  convex  when  the  section 
formed  by  any  plane  intersecting  it  is  a  convex  polygon. 

All  the  polyedrons  treated  of  in  this  work  will  be  under- 
stood to  be  convex. 

4.  Definition.  The  volume  of  any  polyedron  is  the  numer- 
ical measure  of  its  magnitude,  referred  to  some  other  poly- 
edron as  the  unit.  The  polyedron  adopted  as  the  unit  is 
called  the  unit  of  volume. 

To  measure  the  volume  of  a  polyedron  is,  then,  to  find  its 
ratio  to  the  unit  of  volume. 
224 


BOOK  VII.  225 

The  most  convenient  unit  of  volume  is  the  cube  whose 
edge  is  the  linear  unit. 

5.  Definition.  Equivalent  solids  are  those  which  have  equal 
volumes. 

PEISMS  AND  PARALLEL0PIPED8. 

6.  Definitions.  A  prism  is  a  polyedron  two  of  whose  oppo- 
site faces,  called  hases^  are  in  parallel  planes, 

and  whose  lateral  edges  (that  is,  the  edges 
intersecting  the  bases)  are  all  parallel  to  the 
same  line. 

From  this  definition  werreadily  deduce  the 
following  consequences  : 

1st.  Any  two  lateral  edges  of  a  prism  are 
parallel  (VI.,  Proposition  Y.,  Corollary). 

2d.  All  the  lateral  faces  of  a  prism  are  parallelograms  (YI., 
Proposition  YIII.).     Hence  all  the  lateral  edges  are  equal. 

3d.  The  bases  of  a  prism  are  equal  polygons  (YI.,  Proposi- 
tion X.). 

The  lateral  faces  of'  a  prism  constitute  its  lateral  or  convex 
surface. 

The  altitude  of  a  prism  is  the  perpendicular  distance  be- 
tween the  planes  of  its  bases  (YI.,  Proposition  IX.) 

A  triangular  prism  is  one  whose  base  is  a  triangle ;  a  quad- 
rangular prism,  one  whose  base  is  a  quadrilateral ;  etc. 

7.  Definitions.  A  right  prism  is  one  whose  lateral 
edges  are  perpendicular  to  the  planes  of  its  faces 
(YI.,  Proposition  XI.) 

In  a  right  prism,  any  lateral  edge  is  equal  to 
the  altitude. 

The  lateral  faces  of  a  right  prism  are  perpen- 
dicular to  the  bases  (YI.,  Proposition  XIY.) 


226 


ELEMENTS  OF   GEOMETRY. 


An  oblique  prism  is  one  whose  lateral  edges  are  oblique  to 
the  planes  of  its  bases. 

In  an  oblique  prism,  a  lateral  edge  is  greater  than  the  altitude. 

8.  Definition.  A  regular  prism  is  a  right  prism  whose  bases 
are  regular  polygons. 


8.  Definition.  Ifaprism,A5CZ)jE^-JF; 
is  intersected  by  a  plane  GK,  not 
pj^rallel  to  its  base,  the  portion  of 
the  prism  included  between  the  base 
and  this  plane,  namely,  \A 5  (7D.&- 
QHIKLy  is  called  a  truncated  prism. 


10.  Definition.  A  right  section  of  a  prism  is  the  section  made 
by  a  plane  passed  through  the  prism  perpendicular  to  one  of 
its  lateral  edges. 

A  right  section  is  perpendicular  to  all  the  lateral  edges 
(YI.,  Proposition  XI.)  and  to  all  the  lateral  faces  (YI.,  Propo- 
sition XIY.)  of  the  prism. 

11.  Definition.  A  parallelopiped  is  a  prism 
whose  bases  are  parallelograms.  It  is  there- 
fore a  polyedron  all  of  whose  faces  are  par- 
allelograms. 

From  this  definition  and  YI.,  Proposi- 
tion X.,  it  is  evident  that  any  two  oppo- 
site faces  of  a  parallelopiped  are  equal  parallelograms. 

12.  Definition.  A  right  parallelopiped  is  a  par- 
allelopiped whose  lateral  edges  are  perpendic- 
ular to  the  planes  of  its  bases.  Hence,  by  YI., 
6,  its  lateral  faces  are  rectangles ;  but  its  bases 
may  be  either  rhomboids  or  rectangles. 


BOOK   VII. 


227 


A  rectangular  parallelopiped  is  a  right  parallelepiped  whose 
bases  are  rectangles.  Hence  it  is  a  parallelopiped  all  of 
whose  faces  are  rectangles. 


13.  Definition.  A  cube  is  a  rectangular  parallelo- 
piped whose  edges  are  all  equal.  Hence  its  faces 
are  all  squares. 


PROPOSITION  I.— THEOREM. 

14.  The  sections  of  a  prism  made  by  parallel  planes  are  equal 
polygons. 


For  the  portion  of  the  prism  in- 
cluded between  the  two  sections  is  a 
new  prism  (6).  Therefore  its  bases, 
which  are  the  sections  in  question, 
are  equal. 

15.  Corollary.  Any  section  of  a 
prism  made  by  a  plane  parallel  to  the 
base  is  equal  to  the  base. 


EXERCISE. 


Theorem. — In  a  rectangular  parallelopiped,  the  four  diagonals 
are  equal  to  each  other ;  and  the  square  of  a  diagonal  is  equal 
to  the  sum  of  the  squares  of  the  three  edges  which  meet  at  a 
common  vertex. 


228 


ELEMENTS   OF   GEOMETRY. 


PROPOSITION  II.— THEOREM. 

16.  The  lateral  area  of  a  prism  is  equal  to  the  product  of  the 
perimeter  of  a  right  section  of  the  prism  by  a  lateral  edge. 

Let  AD'  be  a  prism,  and  GHIKL  a 
right  section  of  it;  then  the  area  of 
the  convex  surface  of  the  prism  is  equal 
to  the  perimeter  GHIKL  multiplied  by 
a  lateral  edge  AA'. 

For,  the  sides  of  the  section  GHIKL^ 
being  perpendicular  to  the  lateral  edges 
AA',  BB'  (YI.,  6),  etc.,  are  the  altitudes 
of  the  parallelograms  which  form  the 
convex  surface  of  the  prism,  if  we  take  as  the  bases  of  these 
parallelograms  the  lateral  edges  AA\  BB\  etc.,  which  are  all 
equal  (6).  Hence  the  area  of  the  sum  of  these  parallelo- 
grams is 

GH  X  AA'  -\-HIX  BB'  +  etc. 
=  {GH-\-HI+  etc.)  X  AA'. 

17.  Corollary.  The  lateral  area  of  a  right  prism  is  equal 
to  the  product  of  the  perimeter  of  its  base  by  its  altitude. 

PROPOSITION  III.— THEOREM. 

18.  Two  prisma  are  equal,  if  three  faces  including  a  triedral 
angle  of  the  one  are  respectively  equal  to  three  faces  similarly 
'placed  including  a  triedral  angle  of  the  other,— 

Let  the  triedral  angles  A 
and  a  of  the  prisms  ABCDE- 
A\  abcde-a',  be  contained  by- 
equal  faces  similarly  placed, 
namely,  ABCDE  equal  to 
abcde,  AB'  equal  to  ab',  and 
AE'  equal  to  aef ;  then  the 
prisms  are  equal. 


BOOK  VII.  229 

For,  superpose  the  second  prism  upon  the  first,  making  the 
base  ahcde  coincide  with  the  equal  base  ABCDE.  Since  the 
diedral  angles  ah  and  AB  are  equal  and  ae  and  AE  are  equal 
(YI.,  Proposition  XXII.),  the  plane  ah'  will  coincide  with  the 
plane  AB'^  and  the  plane  ae'  with  the  plane  AE',  Hence  the 
intersection  aa'  will  fall  along  the  intersection  AA',  As  the 
faces  ah'  and  AB'  are  equal,  and  have  now  been  suitably 
superposed,  they  must  coincide  throughout^  and  a'h'  will 
coincide  with  A'B'.  For  the  same  reason,^  a'e'  will  coincide 
with  A'E'.  Consequently,  the  plane  determined  by  a'h'  and 
aV,  namely,  the  plane  of  the  upper  base  of  the  second  prism, 
will  coincide  with  the  plane  of  the  upper  base  of  the  first 
prism.  Any  lateral  edge,  as  eef^  will  fall  along  the  corre- 
sponding lateral  edge  EE'^  for  they  are  now  parallel  to  the 
same  line  AA'^  and  have  a  point  e  of  one  coinciding  with  a 
point  E  of  the  other.  They  have  thus  the  same  direction 
^  and  a  point  in  common,  and  must  coincide  throughout^  by  I., 
Postulate  II. 

Since  all  the  lateral  edges  of  the  second  prism  coincide 
with  the  corresponding  lateral  edges  of  the  first,,  the  planes 
of  all  the  corresponding  lateral  faces  must  coincide.  There- 
fore, as  all  the  corresponding  faces  of  the  two  prisms  coincide 
(the  bases  included),  the  prisms  are  equal. 

19.  Corollary  I.  Two  truncated  prisms  are  equal,  if  three 
faces  including  a  triedral  angle  of  the  one  are  respectively  equal 
to  three  faces  similarly  placed  including  a  triedral  angle  of  the 
other.  For  the  preceding  demonstration  applies  whether  the 
planes  A'B'C'D'E'  and  a'h'c'dle  are  parallel  or  inclined  to  the 
lower  bases. 

20.  Corollary  II.  Two  right  prisms  are  equals  if  they  have 
equal  bases  and  equal  altitudes. 


20 


230 


ELEMENTS   OF  GEOMETRY. 


/ 


In  the  case  of  right  prisms,  it  is  not  necessary  to  add  the 
condition  that  the  faces  shall  be 
similarly  placed ;  for  if  the  two 
right  prisms  ABC-A! ^  dbc-a\  can- 
not be  made  to  coincide  by  placing 
the  base  ABC  upon  the  equal  base 
ahc ;  yetj  by  inverting  one  of  the 
prisms  and  applying  the  base 
ABC  to  the  base  a'6V,  they  will 
coincide. 

PROPOSITION  IV.— THEOREM. 

V  21.  Any  oblique  prism  is  equivalent  to  a  right  prism  whose 
base  is  a  right  section  of  the  oblique  prism,  and  whose  altitude  is 
equal  to  a  lateral  edge  of  the  oblique  prism. 

JjQtABCDE-A^hG  the  oblique  prism. 
At  any  point  F  in  the  edge  AA\  pass 
a  plane  perpendicular  to  AA'  and 
forming  the  right  section  FGHIK. 
Produce  AA'  to  F\  making  FF'  z= 
AA\  and  through  F'  pass  a  second 
plane  perpendicular  to  the  edge  AA', 
intersecting  all  the  faces  of  the  prism 
produced,  and  forming  another  right 
section  F'G'H'I'K'  parallel  and  equal 

to  the  first.  The  prism  FGHIK-F'  is  a  right  prism  whose 
base  is  the  right  section  and  whose  altitude  FF'  is  equal  to 
the  lateral  edge  of  the  oblique  prism. 

The  solid  ABCDE-F  is  a  truncated  prism  which  is  easily 
shown  to  be  equal  to  the  truncated  prism  A'B'C'D'E'-F' 
(Proposition  III.,  Corollary  I.).  Taking  the  first  away  from 
the  whole  solid  ABCDE-F',  there  remains  the  right  prism ; 


BOOK    VII. 


231 


taking  the  second  away  from  the  same  solid,  there  remains 
the  oblique  prism ;  therefore  the  right  prism  and  the  oblique 
prism  have  the  same  volume ;  that  is,  they  are  equivalent. 

^V^..^^      PROPOSITION  v.— THEOREM. 

26.  Any  parallelopiped  is  equivalent  to  a  rectangular  parallel- 
opiped  of  the  same  altitude  and  an  equivalent  base. 

Let  ABGD^A'  be 
any  oblique  parallel- 
opiped whose  base  is 
A  BCD  J  and  altitude 
B'O. 

Produce  the  edges 
AB,A'B',I)C,D'C'; 
in  AB  produced 
take  FG  =  AB,  and 
through   F  and    G 

pass  planes  FF'FI,  GG'H'H,  perpendicular  to  the  produced 
edges ;  then  the  given  parallelopiped  and  the  right  parallelo- 
piped FF'FI-H  are  equivalent,  by  Proposition  lY. 

Produce,  now,  the  edges  of  this  second  parallelopiped  ZF, 
FF\HG,  H'G';  in  IF  produced  take  NK^IF,  and  through 
.N  and  K  pass  planes  KLL'K'  smd  NMM'N'  perpendicular 
to  the  produced  edges.  Then  the  second  parallelopiped  and 
the  parallelopiped  NMM'W-K  are  equivalent,  by  Proposi- 
tion lY.  Consequently,  the  given  parallelopiped  and  the 
parallelopiped  NMM'N'-K  are  equivalent.  The  last-named 
parallelopiped  is  a  right  parallelopiped,  by  construction,  since 
the  face  KLL'K'  was  drawn  perpendicular  to  the  lateral 
edges.  Moreover,  as  the  planes  KL'  and  KN'  are  perpendic- 
ular, the  first  to  KI,  the  second  to  AG,  they  are  perpendicular 
to  the  plane  AHK,  by  YI.,  Proposition  XIY.,  and  their  inter- 


232 


ELEMENTS   OF   GEOMETRY. 


section  KK'  is  perpendicular  to  AHK(Y1.^  Proposition  XYI.)> 
and  therefore  to  KL 

(YL,  6).    Hence  the  ^  ^'  ''  ^' 

base  KLL'K'  is  a 
rectangle,  and  the 
par  allelopiped 
Xii'X'-iV^is  a  rec- 
tangular parallelopi- 
ped.  If,  now,  we 
take  KIjMN  as  its 
base,  its   altitude  is 

equal  to  that  of  the  given  parallelopiped,  since  the  planes 
AUK  and  A'H'K'  are  parallel;  and  its  base  is  equivalent 
to  ABGD^  since  each  of  them  is  equivalent  to  FGHI  (lY., 
Proposition  I.). 

^  PROPOSITION  VI.— THEOREM. 

23.  The  plane  passed  through  two  diagonally  opposite  edges 
of  a  parallelopiped  divides  it  into  two  equivalent  triangular 
prisms. 

Let  ABCD-A'  be  any  parallelopi- 
ped ;  the  plane  A  CCA',  passed  through 
its  opposite  edges  AA'  and  CC,  divides 
it  into  two  equivalent  triangular  prisms 
ABC- A'  and  ADC-A\ 

Let  FGHI  be  any  right  section  of 
the  parallelopiped,  made  by  a  plane 
perpendicular  to  the  edge  A  A'.     The 

intersection,  FH,  of  this  plane  with  the  plane  AC  is  the 
diagonal  of  the  parallelogram  FGHI,  and  divides  that  par- 
allelogram into  two  equal  triangles,  FGH  and  FIH.  The 
oblique  prism  ABG-A'  is  equivalent  to  a  right  prism  whose 


BOOK    VII. 


233 


base  is  the  triangle  FGH  and  whose  altitude  is  AA'  (Propo- 
sition lY.);  and  the  oblique  prism  ADC- A'  is  equivalent  to 
a  right  prism  whose  base  is  the  triangle  FIH  and  whose  alti- 
tude is  AA\  The  two  right  prisms  are  equal  (Proposition 
III.,  Corollary  II.) ;  therefore  the  oblique  prisms,  which  are 
respectively  equivalent  to  them,  are  equivalent  to  each  other. 

i-  PKOPOSITION  VII.— THEOREM. 

24.  Tioo  rectangular  parallelopipeds  having  equal  bases  are  to 
each  other  as  their  altitudes. 

Let  P  and  Q  be  two  rectangular 
parallelopipeds  having  equal  bases, 
and  let  AB  and  CD  be  their  alti- 
tudes. 

Ist.  Suppose  the  altitudes  have 
a  common  measure,  which  is  con- 
tained m  times  in  AB  and  n  times 
in  CD.    Then  we  have 

AB  ^m 
CD~  n' 

Apply  this  measure  to  AB  and  CJ),  and  through  the  poinds 
of  division  draw  planes  perpendicular  to  AB  and  CD.  P 
will  thus  be  divided  into  m,  and  Q  into  n,  smaller  parallelo- 
pipeds, all  of  which  will  be  equal,  by  Proposition  I.,  Corol- 
lary, and  Proposition  III.,  Corollary  II.    Hence 


p 

\ N 

\ !\ 

\ K 

\ I\ 

\ \ 

Therefore 


P 

Q 


m 

n' 


P^AB 
i        Q        CD' 

The  proof  may  be  extended  to  the  case  where  the  altitudes 
are  incommensurable,  by  the  method  exemplified  in  the  proof 

20* 


^/ 


234 


ELEMENTS  OF   GEOMETRY. 


J 


of  II.,  Proposition  XII.,  of  III.,  Proposition  I.,  and  of  lY., 
Proposition  II. 
^'25.  Scholium,  The  three  edges  of  a  rectangular  parallelo- 
$     piped  which  meet  at  a  common  vertex  are  called  its  dimen- 
sions j  and  the  preceding  theorem  may  be  expressed  as  follows : 
Two  rectangular  parallelopipeds  which  have  two  dimensions 
in  common  are  to  each  other  as  their  third  dimensions. 

f  PROPOSITION  VIII.— THEOREM. 

^6.  T.0  rectangular  parallelopi,e,s  Kaun,  e.ual  aUitudes 
are  to  each  other  as  their  bases. 

Let  a,  b,  and  c  be  the  three  dimen- 
sions of  the  rectangular  parallelo- 
piped  P;  m,  w,  and  c  those  of  the 
rectangular  parallelopiped  Q;  the 
dimension  c,  or  the  altitude,  being 
common. 

Construct  B,  sl  third  rectangular 
parallelopiped,  having  the  dimensions 
m,  b,  and  c. 

If  a  and  m  are  taken  as  the  alti- 
tudes of  P  and  JR,  their  bases  are 
equal,  and,  by  Proposition  YII., 

P^  a 
1^      m* 

If  b  and  n  are  taken  as  the  altitudes  of  R  and  §,  theii 
bases  are  equal,  and,  by  Proposition  VII., 

R^b,  . 
Q       n' 

and,  multiplying  these  ratios  together, 

P^  a  X  b     ' 
Q       m  X  n 


p 

F^ 

Q 

rA 

1 
J 

A 

\ 

n' 


BOOK  vir. 


awRft 


235 


But  a  y^  b  is  the  area  of  the  base  of  P,  awFm  X  w  is  the 
area  of  the  base  of  Q;  therefore  P  and  Q  are  in  the  ratio 
of  their  bases. 

27.  Scholium.  This  proposition  may  also  be  expressed  as 
follows : 

Two  rectangular  parallelopipeds  which  have  one  dimension  in 
common,  are  to  each  other  as  the  products  of  the  other  two 
dimensions. 


PROPOSITION  IX.— THEOREM. 

28.  Any  two  rectangular  parallelopipeds  are  to  each  other  as 
the  products  of  their  three  dimensions. 

Let  a,  by  and  c  be  the  three  dimen- 
sions of  the  rectangular  parallelo- 
piped  P;  m,  n,  and  p  those  of  the 
rectangular  parallelopiped  Q. 

Construct  P,  a  third  rectangular 
parallelopiped,  having  the  dimen- 
sions a,  6,  and  p.  By  Proposition 
YII.  we  have 


p 

\      K 

C            1 

L._P_ ! 

P 
P 

c 

and 

by  Proposition 

YIII., 

P 

_  axb  , 

Q 

m  Xn' 

and, 

multiplying  these  ratios  together, 

P 

axb  X  c 

Q 

mXnXp 

\    K 

V 


4 


236  ELEMENTS   OF   GEOMETRY. 

PROPOSITION  X.— THEOREM.      /4^ 

The  volume  of  a  rectangular  parallelopiped  is  equal  to  the 
product  of  its  three  dimensions,  the  unit  of  volume  being  the  cube 
whose  edge  is  the  linear  unit. 

Let  a,  b,  c,  be  the  three  dimen- 
sions of  the  rectangular  parallelo- 
piped P;  and  let  Q  be  the  cube 
whose  edge  is  the  linear  unit.  The 
three  dimensions  of  Q  are  each 
equal  to  unity,  and  we  have,  by 
the  preceding  proposition, 


p 

\    K 

i 

p 
Q 


ax  b  X  c 
1  X  1  X  1 


=  aXbXc. 


P., 


Now,  Q  being  takeA  as  the  unit  of  volume,  -   is  the  numer- 

ical  measure,  or  volume  of  P,  in  terms  of  this  unit  (4); 
therefore  the  volume  of  P  is  equal  to  the  product  ^  X  ^  X  <^- 

30.  Scholium  I.  Since  the  product  a  X  b  represents  the 
base,  when  c  is  called  the  altitude,  of  the  parallelopiped,  this 
proposition  may  also  be  expressed  as  follows : 

The  volume  of  a  rectangular  parallelopiped  is  equal  to  the 
product  of  its  base  by  its  altitude. 

31.  Scholium  II.  When  the  three  dimen- 
sions of  the  parallelopiped  are  each  ex- 
actly divisible  by  the  linear  unit,  the  truth 
of  the  proposition  is  rendered  evident  by 
dividing  the  solid  into  cubes,  each  of  which 
is  equal  to  the  unit  of  volume.  Thus,  if 
the  three  edges  which  meet  at  a  common 

vertex  A  are,  respectively,  equal  to  3,  4,  and  5  times  the  linear 
unit,  these  edges  may  be  divided  respectively  into  3,  4,  and  5 


t\     ^  "^^^ 

V.    ^    v.  ^ 

BOOK  VII.  237 

equal  parts,  and  then  planes  passed  through  the  several  points 
of  division  at  right  angles  to  these  edges  will  divide  the  solid 
into  cubes,  each  equal  to  the  unit  cube,  the  number  of  which 
is  evidently  3x4x5- 

But  the  more  general  demonstration,  above  given,  includes 
also  the  cases  in  which  one  of  the  dimensions,  or  two  of 
them,  or  all  three,  are  incommensurable  with  the  linear  unit. 

32.  Scholium  III.  If  the  three  dimensions  of  a  rectangular 
parallelopiped  are  each  equal  to  <2,  the  solid  is  a  cube  whose 
edge  is  a,  and  its  volume  is  a  X  <^  X  <^  =  <^' ;  or,  the  volume 
of  a  cube  is  the  third  power  of  its  edge.  Hence  it  is  that  in 
arithmetic  and  algebra  the  expression  "  cube  of  a  number" 
has  been  adopted  to  signify  the  "  third  power  of  a  number." 

PROPOSITION  XI.— THEOREM. 

33.  The  volume  of  any  parallelopiped  is  equal  to  the  product 
of  the  area  of  its  base  by  its  altitude. 

For,  by  Proposition  Y.,  the  volume  of  any  parallelopiped  is 
equal  to  that  of  a  rectangular  parallelopiped  having  an  equiv- 
alent base  and  the  same  altitude  (30). 

PROPOSITION  XII.— THEOREM.        ' 

34.  The  volume  of  a  triangular  prism  is  equal  to  the  product 
of  its  base  by  its  altitude. 

Let  ABC-A'  be  a  triangular  prism. 
In  the  plane  of  the  base  complete  the 
parallelogram  ABGD,  and  then  through 
D  draw  a  line  DD'  parallel  to  AA\  and 
through  DD'  and  CO',  and  DD'  and 
AA\  pass  planes,  thus  constructing  the 


n/ 


238 


ELEMENTS   OF   GEOMETRY. 


parallelopiped  ABCD-D'.    The  given  prism  is  half  of  the 
parallelepiped,  by  Proposition  YI.,  and 
it  has  the  same  altitude. 

The  volume  of  the  parallelopiped  is 
equal  to  its  base  BD  multiplied  by  its 
altitude  (Proposition  XI.) ;  therefore 
the  volume  of  the  triangular  prism  is 
equal  to  its  base  ABC,  the  half  of  BD, 
multiplied  by  its  altitude. 

.     35.  Corollary.  The  volume  of  any  prism  is  equal  to  the 
product  of  its  base  by  its  altitude. 

Let  ABCDE-A'  be  any  prism.  It  may 
be  divided  into  triangular  prisms  by  planes 
passed  through  a  lateral  edge  AA'  and  the 
several  diagonals  of  its  base.  The  volume 
of  the  given  prism  is  the  sum  of  the  vol- 
umes of  the  triangular  prisms,  or  the  sum 
of  their  bases  multiplied  by  their  common  ''- 

altitude,  which  is  the  base  ABODE  of  the  given  prism  multi- 
plied by  its  altitude. 


PYKAMIDS. 

36.  Definitions.  A  pyramid  is  a  polyedron  bounded  by  a 
polygon  and  triangular  faces  formed  by 
the  intersections  of  planes  passed  through  b 

the  sides  of  the  polygon  and  a  common 
point  out  of  its  plane  ;  as  S-ABGDE. 

The  polygon,  ABODE,  is  the  base  of 
the  pyramid;  the  point,  S,  in  which  the 
triangular  faces  meet,  is  its  vertex;  the 
triangular  faces  taken  together  constitute 
its  lateral,  or  convex,  surface;  the  area  of 


BOOK  VII.  239 

this  surface  is  the  lateral  area;  the  lines  SA^  SB^  etc.,  in 
which  the  lateral  faces  intersect,  are  its  lateral  edges.  The 
altitude  of  the  pyramid  is  the  perpendicular  distance  SO  from 
the  vertex  to  the  base. 

A  triangular  pyramid  is  one  whose  base  is  a  triangle;  a 
quadrangular  pyramid,  one  whose  base  is  a  quadrilateral ;  etc. 

A  triangular  pyramid,  having  but  four  faces  (all  of  which 
are  triangles),  is  a  tetraedron ;  and  any  one  of  its  faces  may 
be  taken  as  its  base. 

37.  Definitions.  A  regular  pyramid  is  one  whose  base  is  a 
regular  polygon,  and  whose  vertex   is   in  the 
perpendicular  to  the  base  erected  at  the  centre  A 
of  the  polygon.     This  perpendicular  is  called 
the  axis  of  the  regular  pyramid. 

From  this  definition  it  can  be  readily  shown 
that  the  lateral  edges  of  a  regular  pyramid  are        / 
all  equal,  and  hence  that  the  lateral  faces  are      /'' 
equal  isosceles  triangles,  ^"^L^^ 

The  slant  height  of  a  regular  pyramid  is  the 
perpendicular  from  the  vertex  to  the  base  of  any  one  of  its 
lateral  faces.     It  is  the  common  altitude  of  all  the  lateral 
faces. 

38.  Definitions.  A  truncated  pyramid  is  the  portion  of  a 
pyramid  included  between  its  base  and  a  plane  cutting  all  its 
lateral  edges. 

When  the  cutting  plane  is  parallel  to  the  base,  the  trun- 
cated pyramid  is  called  a  frustum  of  a  pyramid.  The  altitude 
of  a  frustum  is  the  perpendicular  distance  between  its  bases. 

In  a  frustum  of  a  regular  pyramid,  the  lateral  faces  are 
equal  trapezoids;  and  the  perpendicular  distance  between 
the  parallel  sides  of  any  one  of  these  trapezoids  is  the  slant 
height  of  the  frustum. 


240 


ELEMENTS   OF   GEOMETRY. 


V 


PEOPOSITION  XIII.— THEOREM. 


39.  If  a  pyramid  is  cut  by  a  plane  parallel  to  its  base :  1st, 
the  edges  and  the  altitude  are  divided  proportionally ;  2d,  the 
section  is  a  polygon  similar  to  the  base. 

Let  the  pyramid  S-ABCDE,  whose  alti- 
tude is  SOj  be  cut  by  the  plane  abode  par- 
allel to  the  base,  intersecting  the  lateral 
edges  in  the  points  a,  6,  c,  d^  e,  and  the  alti- 
tude in  0  ;  then 

1st.  The  edges  and  the  altitude  are  dir 
voided  proportionally. 

Pass  a  plane  through  the  altitude  SO 
and  any  lateral  edge  SA,  cutting  the  base 
in  AO  and  the   section  in  ao.     By  YI., 
Proposition  YIII.,  _,ab,  be,  cd,  . . .  ao  are  parallel  respectively 
to  AB,  BC,  CD,  ...AO.    Therefore,  by  III.,  Proposition  I., 


8a 

Sb 

Sc 

Sd 

So 

SA 

SB 

SG 

SD" 

'       SO' 

2d.  The  section  abode  and  the  base  are  similar.  For  they 
are  mutually  equiangular,  by  YI.,  Proposition  X.,  and  by 
similar  triangles  we  have 


ab  Sa       be  Sb       cd  Sc 

AB  ~'SA'    BC~  'SB'     CD  ~  SC 


whence 


ab 
AB 


bc_ 
BC 


cd 
CD 


and  the  homologous  sides  of  the  polygons  are  proportional. 
Therefore  the  section  and  the  base  are  similar. 


BOOK   VII. 


241 


40.  Corollary  1.  If  a  pyramid  is  cut  by  a  plane  parallel  to 
its  base,  the  area  of  the  section  is  to  the  area  of  the  base  as  the 
square  of  its  distance  from  the  vertex  is  to  the  square  of  the 
altitude  of  the  pyramid. 

For 

abcde    ^  a^  ^^  jy^  Proposition  IX. ; 


ABODE 

Z3^'   -^       '     "^ 

but 

ab   ,      Sa        So 

AB~~  SA~  SO 

Therefore 

abcde          So 

ABCDE      SO' 

41.  Corollary  II.  If  two  pyramids  have  equal  altitudes  and 
equivalent  bases,  sections  made  by  planes  parallel  to  their  bases 
and  at  equal  distances  from  their  vertices  are  equivalent. 


PROPOSITION  XIV.— THEOREM. 

42.  The  lateral  area  of  a  regular  pyramid  is  equal  to  the 
product  of  the  perimeter  of  its  base  by  one-half  its  slant 
height. 

For,  let  S- ABCDE  be  a  regular  pyr- 
amid ;  the  lateral  faces  SAB,  SBC,  etc., 
being  equal  isosceles  triangles,  whose 
bases  are  the  sides  of  the  regular  poly- 
gon ABCDE  and  whose  common  alti- 
tude is  the  slant  height  SH,  the  sum 
of  their  areas,  or  the  lateral  area  of 
the  pyramid,  is  equal  to  the  sum  of 
AB,  BC,  etc.,  multiplied  by  \SH. 

L       g  21 


V 


242 


ELEMENTS   OF   GEOMETRY. 


43.  Corollary.  The  lateral  area  of  the  frustum  of  a  regular 
pyramid  is  equal  to  the  half  sum  of  the  perimeters  of  its  bases 
multiplied  by  the  slant  height  of  the  frustum. 


PROPOSITION  XV.— THEOREM. 

44.  If  the  altitude  of  any  given  triangular  pyramid  is  divided 
into  equal  parts,  and  through  the  points  of  division  planes  are 
passed  parallel  to  the  base  of  the  pyramid,  and  on  the  sections 
made  by  these  planes  as  upper  bases  prisms  are  described  having 
their  edges  parallel  to  an  edge  of  the  pyramid  and  their  altitudes 
equal  to  one  of  the  equal  parts  into  which  the  altitude  of  the 
pyramid  is  divided,  the  total  volume  of  these  prisms  will  approach 
the  volume  of  the  pyramid  as  its  limit  as  the  number  of  parts 
into  which  the  altitude  of  the  pyramid  is  divided  is  indefinitely 
increased. 

Let  S-ABC  be  the  given  trian- 
gular pyramid,  whose  altitude  is 
A  T.  Divide  the  altitude  A  T  into 
any  number  of  equal  parts  Ax, 
xy,  etc.,  and  denote  one  of  these 
parts  by  h.  Through  the  points 
of  division  x,  y,  etc.,  pass  planes 
parallel  to  the  base,  cutting  from 
the  pyramid  the  sections  DBF, 
GUI,  etc.      Upon   the   triangles 

DEF,  GHI,  etc.,  as  upper  bases,  construct  prisms  whose 
lateral  edges  are  parallel  to  SA,  and  whose  altitudes  are  each 
equal  to  h.  This  is  effected  by  passing  planes  through  EF, 
HI,  etc.,  parallel  to  SA.  There  will  thus  be  formed  a  series 
of  prisms  DEF- A,  GSI-Dj^tc,  inscribed  in  the  pyramid. 


/(■^  rror^   tW 


IvI^A  ' 


BOOK  VII.  243 

Again,  upon  the  triangles  ABC^  DEF,  GHI^  etc.,  as  lower 
bases,  construct  prisms  whose  lateral  edges  are  parallel  to 
SAj  and  whose  altitudes  are  each  equal  to  h.  This  also  is 
effected  by  passing  planes  through  BG,  JSF,  HI^  etc.,  parallel 
to  SA.  There  will  thus  be  formed  a  series  of  prisms  ABC-D, 
DEF-G,  etc.,  which  may  be  said  to  be  circumscribed  about 
the  pyramid. 

The  total  volume  of  the  inscribed  prisms  is  obviously  less 
and  the  total  volume  of  the  circumscribed  prisms  is  obviously 
greater  than  the  volume  of  the  pyramid. 

Each  inscribed  prism  is  equivalent  to  the  circumscribed 
prism  immediately  above  it,  since  they  have  the  same  base 
and  equal  altitudes.  Consequently,  the  difference  between 
the  total  volume  of  the  inscribed  prisms  and  the  total  volume 
of  the  circumscribed  prisms  is  the  volume  of  the  lowest  cir- 
cumscribed prism  ABC-D,  and  therefore  the  difference  be- 
tween the  total  volume  of  the  inscribed  prisms  and  the 
volume  of  thp  pyramid  is  less  than  the  volume  of  ABC-D. 
"By  increasing  at  pleasure  the  number  of  parts  into  which 
the  altitude  AT  is  divided,  we  can  make  the  volume  of 
ABC-D  as  small  as  we  please,  since  we  diminish  its  altitude 
at  pleasure  without  changing  its  base.  Therefore  we  can 
make  the  difference  between  the  total  volume  of  the  in- 
scribed prisms  and  the  volume  of  the  pyramid  as  small  as 
we  please ;  but,  as  we  have  seen  above,  we  cannot  make  it 
absolutely  zero.  Hence  the  volume  of  the  pyramid  is  the 
limit  of  the  total  volume  of  the  inscribed  prisms,  as  the 
number  of  parts  into  which  the  altitude  AT  ia  divided  is 
indefinitely  increased. 


244  ELEMENTS   OF   GEOMETRY. 

PROPOSITION  XVI.— THEOREM. 

45.  Two  triangular  pyramids  having  equivalent  bases  and 
equal  altitudes  are  equivalent. 

Let  S-ABC  and  8'-A'B^C'  be  two  triangular  pyramids 


G        A! 


having  equivalent  bases  ABC^  A'B'C'^  in  the  same  plane,  and 
a  common  altitude  A  T. 

Divide  the  altitude  A  T  into  any  arbitrarily  chosen  number 
n  of  equal  parts,  Ax^  xy^  yz^  etc.,  and  through  the  points  of 
division  pass  planes  parallel  to  the  plane  of  the  bases,  inter- 
secting the  two  pyramids.  In  the  pyramid  S-ABG  inscribe 
a  series  of  prisms  whose  upper  bases  are  the  sections  BEF, 
GHIj  etc.,  and  in  the  pyramid  S'-A'B'C  inscribe  a  series  of 
prisms  whose  upper  bases  are  the  sections  D'B'F'.,  G'HT, 
etc.  Since  the  corresponding  sections  are  equivalent  (Propo- 
sition XIII.,  Corollary  II.),  the  corresponding  prisms,  having 
equivalent  bases  and  equal  altitudes,  are  equivalent;  there- 
fore the  sum  of  the  prisms  inscribed  in  the  pyramid  S-ABG 
is  equivalent  to  the  sum  of  the  prisms  inscribed  in  the 
pyramid  S'-A'B'C;  that  is,  if  we  denote  the  total  volumes 
of  the  two  series  of  prisms  by  V  and  F',  we  have 


BOOK   VII. 


245 


no  matter  what  the  value  of  n.    If  we  vary  w,  V  and  F' 
obviously  vary. 

If  n  is  indefinitely  increased,  V  has  the  volume  of  the  pyr- 
amid S-ABG,  and  F'  the  volume  of  the  pyramid  S'-A'B'C, 
as  its  limit  (Proposition  XY.).  Therefore,  by  III.,  Theorem 
of  Limits,  these  volumes  are  equal. 

PROPOSITION  XVII.— THEOREM. 
i|fp  A  triangular  pyramid  is  one-third  of  a  triangular  prism 
of  the  same  base  and  altitude. 

Let  S-ABC  be  a  triangular  pyramid.  Through  A  and  C 
draw  the  lines  AE  and  CD  parallel  to  BS, 
Through  AE  and  CD,  which  are  parallel,  by 
VI.,  Proposition  V.,  Corollary,  pass  a  plane,  and 
through  S  pass  a  second  plane  parallel  to  ABC. 
The  prism  ABC-E  has  the  same  base  and  alti- 
tude as  the  given  pyramid,  and  we  are  to  prove 
that  the  pyramid  is  one-third  of  the  prism. 

Taking  away  the  pyramid  S-ABC  from  the 
prism,  there  remains  a  quadrangular  pyramid  whose  base  is 
the  parallelogram  ACDE  and  vertex  S.  The  plane  SEC^ 
passed  through  SE  and  SC^  divides  this  pjrramid  into  two 
triangular  pyramids,  S-AEC  and  S-ECD,  which  are  equiva- 
lent to  each  other,  since  their  triangular  bases  AEC  and  ECD 
are  the  halves  of  the  parallelogram  A  CDEj  and  their  common 
altitude  is  the  perpendicular  from  S  upon  the  plane  ACDE 
(Proposition  XVI.).  The  pyramid  S-ECD  may  be  regarded 
as  having  ESD  as  its  base  and  its  vertex  at  C;  therefore  it 
is  equivalent  to  the  pyramid  S-ABC,  which  has  an  equivalent 
base  and  the  same  altitude.  Therefore  the  three  pyramids 
into  which  the  prism  is  divided  are  equivalent  to  each  other, 
and  the  given  pyramid  is  one-third  of  the  prism. 

21*  ; 


V 


t 


246  ELEMENTS  OF  GEOMETRY. 

47.  Corollary.  The  volume  of  a  triangular  pyramid  is  equal 
to  one-third  of  the  product  of  its  base  by  its  altitude. 

^^^  PEOPOSITION  XVIII.— THEOREM. 

48.  The  volume  of  any  pyramid  is  equal  to  one-third  of  the 
product  of  its  base  by  its  altitude. 

For  any  pyramid,  S-ABCDB,  may  be  di- 
vided into  triangular  pyramids  by  passing 
planes  through  an  edge  SA  and  the  diagonals 
ABj  ACj  etc.,  of  its  base.    The  bases  of  these 
pyramids  are  the  triangles  which  compose 
the  base  of  the  given  pyramid,  and  their 
common  altitude  is  the  altitude  SO  of  the 
given  pyramid.    The  volume  of  the  given  ^ 
pyramid  is  equal  to  the  sum  of  the  volumes  of  the  triangular 
pyramids,  which  is  one-third  of  the  sum  of  their  bases  multi- 
plied by  their  common  altitude,  or  one-third  of  the  product 
of  the  base  ABODE  by  tte  altitude  SO. 

49.  Scholium.  The  volume  of  any  polyedron  may  be  found 
by  dividing  it  into  pyramids,  and  computing  the  volumes  of 
these  pyramids  separately.  The  division  may  be  effected  by 
taking  a  point  within  the  polyedron  and  joining  it  with  all 
the  vertices.  The  polyedron  will  then  be  decomposed  into 
pyramids  whose  bases  will  be  the  faces  of  the  polyedron,  and 
whose  common  vertex  will  be  the  point  taken  within  it. 

PROPOSITION  XIX.— THEOREM. 

60.  A  frustum  of  a  triangular  pyramid  is  equivalent  to  the 
sum  of  three  pyramids  whose  common  altitude  is  the  altitude  of 
the  frustum,  and  whose  bases  are  the  lower  base,  the  upper  base, 
and  a  mean  proportional  between  the  bases,  of  the  frustum. 


1  '^  ' 

1V 

BOOK  vir.  ;,';  247 

Let  ABC-D  be  a  frustum  of  a  triangular  pyramid,  the 
plane  DEF  being  parallel  to  the  base  ABC. 

Through  the  vertices  A,  B,  and  ^ J' 

C  pass  a  plane  ^-E^C,  and  through  //  y^\/  \ 

the  vertices  E,  D,  and  0  pass  a  /y^ '  \y^\  \ 

plane  ^DC,  dividing  the  frustum  /^  I  /   /'  a\ 

into  three  pyramids.  ^C^^/i'J!jf---j53^^'^ 

The  first  of  these,  ABG-E,  has  ^^^l^""'^'^ 

for  its  base  the  lower  base  of  the  (^^^ 

frustum,  and  for  its  altitude  the  altitude  of  the  frustum  ;  the 
second,  DEF-C,  has  for  its  base  the  upper  base  of  the  frus- 
tum, and  for  its  altitude  the  altitude  of  the  frustum.  It 
remains  to  show  that  the  third,  ACD-E,  is  equivalent  to  a 
pyramid  having  for  its  altitude  the  altitude  of  the  frustum, 
and  for  its  base  a  mean  proportional  between  the  bases  of 
the  frustum. 

Through  E  in  the  plane  ABED  draw  a  line  EEf  parallel  to 
AD,  and  through  E',  D,  and  C  pass  a  plane.  EE'  is  parallel 
to  the  plane  ACFD,  by  YI.,  Proposition  YI.  Therefore  the 
pyramids  ACD-E  and  ACD-E'  are  equivalent,  since  they 
have  the  same  base  and  equal  altitudes.  If  we  take  D  as 
the  vertex  and  AE'C  as  the  base  of  ACD-E' ,  it  has  for  its 
altitude  the  altitude  of  the  frustum. 

Through  F  in  the  plane  ACFD  draw  FF'  parallel  to  AD, 
AE'F'-D  is  a  prism,  and  consequently  its  bases  DEF  and 
AE'F'  are  equal^  and  E'F'  is  parallel  to  EFy  and  therefore 

to  5a 

AE'F'  ^  AF' 
AE'C         AC' 

since  the  triangles  AE'F'  and  AE'C  have  the  same  altitude. 

AE'C       AE'   n     .. 

=  -j^-,  for  the  same  reason. 
AJjC        ab 


248 


ELEMENTS   OF   GEOMETRY. 


AF'         AE' 


AG 


AB 


by  III.,  Proposition  I. 
Therefore 

DEF  ^AE'C 
AE'C       ABC' 

and  the  base  of  the  pyramid  AE'C-D  is  a  mean  proportional 
between  the  bases  of  the  frustum. 

^1.  Corollary.  A  frustum  of  any  pyramid  is  equivalent  to 
the  sum  of  three  pyramids  whose  common  altitude  is  the  altitude 
of  the  frustum,  and  whose  bases  are  the  lower  base,  the  upper 
base,  and  a  mean  proportional  between  the  bases,  of  the  frustum. 


Suggestion,  Let  ABCDE-F  be  a  frustum  of  any  pyramid 
S-ABCDE.  Construct  a  triangular  pyramid,  S'-A'B'C, 
having  the  same  altitude  as  S-ABCDE,  and  a  base  A'B'C 
equivalent  to  ABODE  and  in  the  same  plane  with  it.  Let 
the  plane  of  the  upper  base  of  the  given  frustum  be  pro- 
duced to  cut  the  triangular  pyramid  in  F'G'T.  The  upper 
bases  of  the  frustums  are  equivalent,  by  Proposition  XIII.) 


BOOK  VII.  249 

Corollary  II.,  and  the  frustums  themselves  are  equivalent, 
since  the  pyramids  are  equivalent  and  the  pyramids  above 
the  frustums  are  equivalent. 

"^  PROPOSITION  XX.— THEOREM. 

52.  A  truncated  triangular  prism  is  equivalent  to  the  sum  of 
three  pyramids  whose  common  base  is  the  base  of  the  prism^  and 
whose  vertices  are  the  three  vertices  of  the  inclined  section. 

F 

Let  ABC-BEF  be  a  truncated  tri- 
angular prism  whose  base  is  ABC  and 
inclined  section  BEF. 

Pass  the  planes  AEC  and  DEC, 
dividing  the  truncated  prism  into  the 
three  pyramids  E-ABC,  E-ACI),  and 
E-CBF. 

The  first  of  these  pyramids,  E-ABC^  has  the  base  ABC 
and  the  vertex  E. 

The  second  pyramid,  E-A  CD,  is  equivalent  to  the  pyramid 
B-ACB ;  for  they  have  the  same  base  ACB,  and  the  same 
altitude,  since  their  vertices  E  and  B  are  in  the  line  EB 
parallel  to  this  base.  But  the  pyramid  B-ACB  is  the  same 
as  B-ABC ;  that  is,  it  has  the  base  ABC  and  the  vertex  B. 

The  third  pyramid,  E-CBF,  is  equivalent  to  the  pyramid 
B-ACF ;  for  they  have  equivalent  bases  CBF  and  ACF  in 
the  same  plane,  and  also  the  same  altitude,  since  their  ver- 
tices E  and  B  are  in  the  line  EB  parallel  to  that  plane.  But 
the  pyramid  B-ACF  is  the  same  as  F-ABC;  that  is,  it  has 
the  base  ABC  and  the  vertex  F. 

Therefore  the  truncated  prism  is  equivalent  to  three  pyra- 
mids whose  common  base  is  ABC  and  whose  vertices  are  E^ 
D,  and  F, 


250 


ELEMENTS   OF   GEOMETRY. 


THE  KEGULAR  POLYEDRONS. 

53.  Definition.  A  regular  polyedron  is  one  whose  faces  are 
all  equal  regular  polygons  and  whose  polyedral  angles  are  all 
equal  to  each  other. 


PROPOSITION  XXI.— THEOREM. 

64..  Only  five  regular  (convex^  polyedrons  are  possible. 

(^The  faces  of  a  regular  polyedron  must  be  regular  polygons, \ 
and  at  least  three  faces  are  necessary  to  form  a  polyedral  \ 
angle ;  moreover,  the  sum  of  the  face  angles  of  a  polyedral  / 
angle  must  be  less  than  four  right  angles )(YI.,  Proposition 
XXI.). 

1st.  The  simplest  regular  polygon  is  the  equilateral  tri- 
angle, and,  since  each  angle  of  an  equilat- 
eral triangle  is  an  angle  of  60°,  three  equi- 
lateral triangles  «an  be  combined  to  form 
a  polyedral  angle.  It  is  probable,  then, 
that  a  regular  polyedron  can  be  formed 
bounded  by  equilateral  triangles  and  hav- 
ing three  at  each  vertex. 

There  is  such  a  regular  polyedron.  It 
has  four  faces,  and  is  called  the  regular 
tetraedron. 

Since  four  angles  of  60°  are  less  than  four  right  angles, 
four  equilateral  triangles  can  be  combined  to  form  a  polyedral 


BOOK   VII. 


251 


angle.  It  is  probable,  then,  that  a  regu- 
lar polyedron  can  be  formed  bounded  by- 
equilateral  triangles  and  having  four  at 
each  vertex. 

There  is  such  a  regular  polyedron. 
It  has  eight  faces,  and  is  called  the  regu- 
lar octaedron. 

Since  five  angles  of  60°  are  less  than  four  right  angles, 
five   equilateral   triangles   can   be   combined 
to  form  a  polyedral  angle.     It  is  probable,  then,  that  a  reg- 
ular polyedron  can  be  formed  bounded  by 
equilateral  triangles  and  having  five  at  each 
vertex. 

There  is  such  a  regular  polyedron.  It 
has  twenty  faces,  and  is  called  the  regular 
icosaedron. 

No  regular  polyedrons  bounded  by  equi- 
lateral triangles  and  having  more  than  five  at  a  vertex  are 
possible.  For  six  or  more  angles  of  60°  cannot  form  a  poly- 
edral angle. 

2d.  The  next  regular  polygon  to  the  equilateral  triangle, 
in  order  of  simplicity,  is  the  square,  each  of  whose  angles  is 
a  right  angle. 

Three  right  angles  can  be  combined  to  form 
a  polyedral  angle.  It  is  probable,  then,  that  a 
regular  polyedron  can  be  formed  bounded  by 
squares  and  having  three  at  each  vertex. 

There  is  such  a  regular  polyedron.  It  has 
six  faces,  and  is  called  the  cuhe^  or  the  regular  hexaedron. 

No  regular  polyedrons  bounded  by  squares  and  having 
more  than  three  at  a  vertex  are  possible.  For  four  or  more 
right  angles  cannot  form  a  polyedral  angle. 


252  ELEMENT8  OP  GEOMETRY. 

3d.  The  next  regular  polygon  is  the  regular  pentagon,  each 
of  whose  angles  contains  108°  (I.,  Proposition  XXYII.). 

Three  angles  of  108°  each  can  be  combined  to  form  a  poly- 
edral  angle.     It  is  probable,  then,  that  a 
regular  polyedron  can  be  formed  bounded 
by  regular  pentagons  and  having  three 
at  each  vertex. 

There  is  such  a  regular  polyedron. 
It  has  twelve  faces,  and  is  called  the 
regular  dodecaedron. 

!No    regular   polyedrons   bounded  by 
regular  pentagons  and  having  more  than  three  at  a  vertex 
are  possible.     For  four  or  more  angles  of  108°  cannot  be 
combined  to  form  a  polyedral  angle. 

4th.  Each  angle  of  the  regular  hexagon  contains  120°.  No 
regular  polyedron  can  be  formed  bounded  by  hexagons.  For 
three  or  more  angles  of  120°  cannot  be  combined  to  form  a 
polyedral  angle. 

No  regular  polyedron  can  be  formed  bounded  by  reg- 
ular polygons  of  more  than  six  sides.  For  it  follows,  from 
I.,  55,  Exercise,  that  the  greater  the  number  of  sides  in  a 
regular  polygon  the  greater  the  magnitude  of  its  angles,  and 
since,  as  we  have  seen,  the  angles  of  a  hexagon  are  too  great 
to  allow  the  existence  of  a  polyedral  angle  whose  plane  faces 
are  regular  hexagons,  those  of  any  regular  polygon  of  more 
than  six  sides  will  be  too  great. 

Therefore  the  only  possible  regular  polyedrons  are  the  five 
we  have  figured. 

55.  Scholium  I.  It  must  be  observed  that  we  have  not 
attempted  to  prove  that  the  five  regular  polyedrons  are  pos- 
sible. This  can  be  done  by  showing  how  to  construct  them ; 
but  the  investigation  is  difficult  and  tedious. 


BOOK   VII. 


253 


56.  Scholium  II.  The  student  may  derive  some  aid  in  com- 
prehending the  preceding  discussion  of  the  regular  polyedrons 
by  constructing  models  of  them,  which  he  can  do  in  a  very 
simple  manner,  and  at  the  same  time  with  great  accuracy,  as 
follows. 

Draw  on  card-board  the  following  diagrams ;  cut  them  out 
entire,  and  at  the  lines  separating  adjacent  polygons  cut  the 
card-board  half  through ;  the  figures  will  then  readily  bend 
into  the  form  of  the  respective  surfaces,  and  can  be  retained 
in  that  form  by  gluing  the  edges. 


V 


Tetraedron. 


Hexaedron, 

Octaedron. 


V 


Dodccaedron. 


Tcosaedron. 


EXERCISES  ON  BOOK  VIL 


THEOREMS. 

1.  The  volume  of  a  triangular  prism  is  equal  to  the  product  of 
the  area  of  a  lateral  face  by  one-half  the  perpendicular  distance 
of  that  face  from  the  opposite  edge. 

2.  The  lateral  surface  of  a  pyramid  is  greater  than  the  base. 
Suggestion.  Join  the  projection  of  the  vertex  on  the  base  with 

the  corners  of  the  base. 

3.  At  any  point  in  the  base  of  a  regular  pyramid  a  perpendic- 
ular to  the  base  is  erected  which  intersects  the  several  lateral  faces 
of  the  pyramid,  or  these  faces  produced.  Prove  that  the  sum  of 
the  distances  of  the  points  of  intersection  from  the  base  is  con- 
stant. 

Suggestion.  The  distances  in  question  are  proportional  to  the 
distances  of  the  foot  of  the  perpendicular  from  the  sides  of  the 
base,  and  these  distances  have  a  constant  sum.  {v.  V.,  Exer- 
cise 16.) 


4.  Two  tetraedrons  which  have 
a  triedral  angle  of  the  one  equal 
to  a  triedral  angle  of  the  other, 
are  to  each  other  as  the  products 
of  the  three  edges  of  the  equal 
triedral  angles,  {v.  IV.,  19,  Ex- 
ercise.) 


6.  In  a  tetraedron,  the  planes  passed  through  the  three  lateral 
edges  and  the  middle  points  of  the  edges  of  the  base  intersect  in 
a  straight  line. 
254 


BOOK  VII.  255 

Suggestion.  The  intersections  of  the  planes  with  the  base  are 
medial  lines  of  the  base.  Therefore  they  intersect  in  the  line 
joining  the  vertex  with  the  point  of  intersection  of  the  medial 
lines  of  the  base. 

6.  The  lines  joining  each  vertex  of  a  tetraedron  with  the  point 
of  intersection  of  the  medial  lines  of  the  opposite  face  all  meet  in 
a  point,  which  divides  each  line  in  the  ratio  1  : 4. 

Note.  This  point  is  the  centre  of  gravity 
of  the  tetraedron. 

Suggestion.  If  AF  and  DG  are  two  of  / 

the  lines  in  question,,  they  must  intersect,  / 

since  they  both  lie  in  the  plane  passed  /  g 
through  AD  and  the  middle  point  E  of  ^V/ 
the  opposite  edge.    Moreover,  since  EF  M 

=  lED  and  EG  =  lEA  (I.,  Exercise  38), 
GFis,  parallel  to  AD  and  is  equal  to  ^AD.  C 

Whence  HF  =  iHA  and   GH  =  iHD. 

The  lines  through  C  and  B  will  also  each  cut  off  \  of  AF,    Hence 
the  four  lines  have  a  common  intersection. 

7.  The  straight  lines  joining  the  middle  points  of  the  opposite 
edges  of  a  tetraedron  all  pass  through  the  centre  of  gravity  of 
the  tetraedron,  and  are  bisected  by  the  centre  of  gravity,   {v.  III., 

Exercise  7.)  ^ 

8.  The  plane  which  bisects  a  diedral  angle  of  a  ie^raedron 
divides  the  opposite  edge  into  segments  which  are  proportional 
to  the  areas  of  the  adjacent  faces. 

Suggestion.  Consider  the  volumes  of  the  two  parts  into  which 
the  tetraedron  is  divided. 

9.  If  a,  6,  e,  d,  are  the  perpendiculars  from  the  vertices  of  a 
tetraedron  upon  the  opposite  faces,  and  a\  6'',  g\  d^,  the  perpen- 
diculars from  any  point  within  the  tetraedron  upon  the  same 
faces  respectively,  then 

abed 

Suggestion.  Join  the  point  in  question  with  the  vertices  of  the 
tetraedron,  and  compare  the  volumes  of  the  four  tetraedrons  thus 
obtained  with  the  volume  of  the  given  tetraedron. 


256 


ELEMENTS   OF   GEOMETRY. 


10.  The  altitude  of  a  regular  tetraedron  is  equal  to  the  sum  of 
the  four  perpendiculars  let  fall  from  any  point  within  it  upon  the 
four  faces. 

11.  Any  lateral  face  of  a  prism  is  less  than  the  sum  of  the  other 
lateral  faces,     {v.  Proposition  II.) 


PROBLEMS. 

12.  Given  three  indefinite  straight  lines  in  space  which  do  not 
intersect,  to  construct  a  parallelopiped  which  shall  have  three  of 
its  edges  on  these  lines,    {v.  VI.,  Exercise  8.) 

13.  Within  a  given  tetraedron,  to  find  a  point  such  that  planes 
passed  through  this  point  and  the  edges  of  the  tetraedron  shall 
divido^lie  tetraedron  into  four  equivalent  tetraedrons.  {v.  Exer- 
cise 6.) ' 


BOOK  YIIL 

THE    THREE    ROUND    BODIES. 

1.  Op  the  various  solids  bounded  by  curved  surfaces,  but 
three  are  treated  of  in  Elementary  Geometry, — namely,  the 
cylinder^  the  cone^  and  the  sphere^  which  are  called  the  three 

ROUND   BODIES. 

THE  CYLINDER. 

2.  Definitions.  A  cylindrical  surface  is  a  curved  surface  gen- 
erated by  a  moving  straight  line  which  continually  touches  a 
given  curve,  and  in  all  of  its  positions  is  parallel  to  a  given 
fixed  straight  line  not  in  the  plane  of  the  curve. 

Thus,  if  the  straight  line  Aa  moves  so  as  continually  to 
touch  the  given  curve  ABCD,  and  so 
that  in  any  of  its  positions,  as  ^6, 
Cc,  Dd,  etc.,  it  is  parallel  to  a  given 
fixed  straight  line  3im,  the  surface 
ABCDdcba  is  a  cylindrical  surface. 
If  the  moving  line  is  of  indefinite 
length,  a  surface  of  indefinite  extent 
is  generated. 

The  moving  line  is  -called  the  generatrix ;  the  curve  which 
it  touches  is  called  the  directrix.  Any  straight  line  in  the 
surface,  as  Bb^  which  represents  one  of  the  positions  of  the 
generatrix,  is  called  an  element  of  the  surface. 

To  draw  an  element  through  any  given  point  of  a  cylin- 
drical surface,  it  is  sufficient  to  draw  a  line  through  the  point 
parallel  to  the  given  fixed  straight  line,  or  parallel  to  ap 
element  (I.,  Postulate  II.). 

r  .  22*  257 


258 


ELEMENTS   OF   GEOMETRY. 


In  this  general  definition  of  a  cylindrical  surface,  the  direc- 
trix may  be  any  curve  whatever.    Hereafter  we  shall  assume 
it  to  be  a  closed  curve,  and  usually  a 
circle,  as  this  is  the  only  curve  whose 
properties  are  treated  of  in  element- 
ary geometry. 

3.  Definition.  The  solid  Ad  bounded 
by  a  cylindrical  surface  and  two  par- . 
allel  planes,  ABD  and  ahd^  is  called 
a  cylinder ;  its  plane  surfaces  ABD, 

abd,  are  called  its  bases;  the  curved  surface  is  sometimes 
called  its  lateral  surface;  and  the  perpendicular  distance 
between  its  bases  is  its  altitude. 

The  elements  of  a  cylinder  are  all  equal. 

A  cylinder  whose  base  is  a  circle  is  called  a  circular  cylinder. 

4.  Definition.  A  right  cylinder  is  one  whose 
elements  are  perpendicular  to  its  base. 

5.  Definition.  A  right  cylinder  with  a  circular 
base,  as  ABCa,  is  called  a  cylinder  of  revolution, 
because  it  may  be  generated  by  the  revolution 
of  a  rectangle  A  Ooa  about  one  of  its  sides,  Oo, 
as  an  axis ;  the  side  Aa  generating  the  curved  ^ 
surface,  and  the  sides  OA  and  oa  generating 

the  bases.  The  fixed  side  Oo  is  the  axis  of  the  cylinder. 
The  radius  of  the  base  is  called  the  radius  of  the  cylinder. 


m\- 


PROPOSITION  I.— THEOREM. 

6.  Every  section  of  a  cylinder  made  by  a  plane  passing  through 
an  element  is  a  parallelogram. 

Let  Bb  be  an  element  of  the  cylinder  Ac ;  then  the  section 
BbdD,  made  by  a  plane  passed  through  Bb,  is  a  parallelogram. 


BOOK   VIII. 


259 


The  line  Dd  in  which  the  cutting  plane  intersects  the 
curved  surface  a  second  time  is  an  ele- 
ment. For,  if  through  any  point  D  of 
this  intersection  a  straight  line  is  drawn 
parallel  to  Bh^  this  line,  by  the  definition 
of  a  cylindrical  surface,  is  an  element  of 
the  surface,  and  it  must  also  lie  in  the 
plane  Bd ;  therefore  this  element,  being 
common  to  both  surfaces,  is  their  intersection. 

The  lines  BD  and  hd  are  parallel  (VI.,  Proposition  YIII.), 
and  the  elements  Bh  and  Dd  are  parallel ;  therefore  Bd  is  a 
parallelogram. 

7.  Corollary.  Every  section  of  a  right  cylinder  made  by  a 
plane  perpendicular  to  its  base  is  a  rectangle. 

PROPOSITION  II.— THEOREM. 

8.  The  bases  of  a  cylinder  are  equal. 

Let  BD  be  the  straight  line  joining  two  points  of  the 
perimeter  of  the  lower  base,  and  let  a 
plane  passing  through  BD  and  the  ele- 
ment Bb  cut  the  upper  base  in  the  line 
bd ;  then  BD  =  bd  (Proposition  I.). 

Let  A  be  any  third  point  in  the  perim- 
eter of  the  lower  base,  and  Aa  the  corre- 
sponding element.  Through  the  parallels 
Aa  and  Bb  pass  a  plane,  and  through  Aa  *^ 

and  Dd  pass  a  plane.  Then  AB  =  ab  and  AD  =  ad  (Propo- 
sition I.)  ;  and  the  triangles  ABD^  abdj  are  equal.  Therefore, 
if  the  upper  base  be  applied  to  the  lower  base  with  the  line 
bd  in  coincidence  with  its  equal  BD^  the  triangles  will  co- 
incide and  the  point  a  will  fall  upon  A  ;  that  is,  any  point  a 
of  the  upper  base  will  fall  on  the  perimeter  of  the  lower  base, 


260  ELEMENTS  OF  GEOMETRY. 

and  consequently  the  perimeters  will  coincide   throughout. 
Therefore  the  bases  are  equal. 

9.  Corollary  I.  Any  two  parallel  sec-  »w/.^^ / 

tions  of  a  cylindrical  surface  are  equal.  /-^^^^^i^V 

For  these  sections  are  the  bases  of  a  A>-^^-   / 

cylinder.  /  .../^/'^ 


10.  Corollary  II.  All  the  sections  of  a  circular  cylinder 
parallel  to  its  bases  are  equal  circles;  and  the  straight  line 
joining  the  centres  of  the  bases  passes  through  the  centres  of  all 
the  parallel  sections.    This  line  is  called  the  axis  of  the  cylinder. 

Suggestion.  In  the  base  draw  two  diameters,  and  through 
these  diameters  and  elements  of  the  cylinder  pass  planes. 
They  will  cut  all  the  sections  in  diameters,  and  their  line  of 
intersections  will  pass  through  all  the  centres. 

11.  Definition.  A  tangent  plane  to  a  cylinder  is  a  plane  which 
passes  through  an  element  of  the  curved  surface  without  cut- 
ting the  surface.  The  element  through  which  it  passes  is 
called  the  element  of  contact 

THE  CONE. 

12.  Definition.  A  conical  surface  is  a  curved  surface  gener 
ated  by  a  moving  straight  line  which  continually  touches  a 
given  curve,  and  passes  through  a  given  fixed  point  not  in 
the  plane  of  the  curve. 

Thus,  if  the  straight  line  SA  moves  so  as  continually  to 
touch  the  given  curve  ABGD,  and  in  all  its  positions,  SB,  SC, 
SBj  etc.,  passes  through  the  given  fixed  point  aS,  the  surface 
S-ABCD  is  a  conical  surface. 


BOOK   VIII. 


261 


The  moving  line  is  called  the  generatrix ;  the  curve  which 
it  touches  is-  called  the  directrix. 
Any  straight  line  in  the  surface,  as 
SBj  which  represents  one  of  the  posi- 
tions of  the  generatrix,  is  called  an 
element  of  the  surface.  The  point  S 
is  called  the  vertex. 

The  straight  line  joining  any  point 
of  a  conical  surface  with  the  vertex 
is  obviously  an  element. 

If  the  generatrix  is  of  indefinite 
length,  as  ASa^  the  whole  surface  generated  consists  of  two 
symmetrical  portions,   each   of  indefinite   extent,   lying  on 
opposite  sides  of  the  vertex,  as  S-ABGD  and  S-abcd,  which 
are  called  nappes ;  one  the  upper,  the  other  the  lower,  nappe. 

13.  Definition.  The  solid  S-ABCD,  bounded  by  a  conical 
surface  and  a  plane  ABD  cutting  the  surface,  is  called  a  cone ; 
its  plane  surface  ABD  is  its  base,  the  point  ;S^  is  its  vertex,  and 
the  perpendicular  distance  SO  from  the  vertex  to  the  base  is 
its  altitude. 

A  cone  whose  base  is  a  circle  is  called  a  circular  cone.  The 
straight  line  drawn  from  the  vertex  of  a  circular  cone  to  the 
centre  of  its  base  is  the  axis  of  the  cone. 

14.  Definition.  A  right  circular  cone  is  a  cir- 
cular cone  whose  axis  is  perpendicular  to  its 
base,  as  S-ABCD. 

The  right  circular  cone  is  also  called  a  cone 
of  revolution,  because  it  may  be  generated  by 
the  revolution  of  a  right  triangle,  SA  0,  about 
one  of  its  perpendicular  sides,  SO,  as  an  axis ; 
the  hypotenuse  SA  generating  the  curved  surface,  and  the 
remaining  perpendicular  side  OA  generating  the  base. 


262 


ELEMENTS   OF   GEOMETRY. 


PROPOSITION  III.— THEOREM. 

15.  Every  section  of  a  cone  made  by  a  plane  passing  through 
its  vertex  is  a  triangle. 

Let  the  cone  S-ABCD  be  cut  by  a 
plane  SBC^  which  passes  through  the 
vertex  S  and  cuts  the  base  in  the  straight 
line  BC ;  then  the  section  SBC  is  a  tri- 
angle; that  is,  the  intersections  SB  and 
SG  with  the  curved  surface  are  straight 
lines. 

For  the  straight  lines  joining  S  with  B  and  C  are  elements 
of  the  surface,  by  the  definition  of  a  cone,  and  they  also  lie 
in  the  cutting  plane ;  therefore  they  coincide  with  the  inter- 
sections of  that  plane  with  the  curved  surface ;  and  BG^  being 
the  intersection  of  two  planes,  is  a  straight  line. 


PROPOSITION  IV.— THEOREM. 

16.  If  the  base  of  a  cone  is  a  circle^  every  section  made  ty  a 
plane  parallel  to  the  base  is  a  circle. 

Let  the  section  abc,  of  the  circular 
cone  S-ABGj  be  parallel  to  the  base. 

Let  0  be  the  centre  of  the  base,  and 
let  0  be  the  point  in  which  the  axis  SO 
cuts  the  plane  of  the  parallel  section. 
Through  SO  and  any  number  of  ele- 
ments SAj  SBj  etc.,  pass  planes  cutting 
the  base  in  the  radii  OA^  OB,  etc.,  and 
the  parallel  section  in  the  straight  lines  oa,  ob,  etc. 
is  parallel  to  OA,  and  ob  to  OB,  we  have 


Since  oa 


ofi         So  „„j   ob         So    „i^  ^     oa         ob 


BOOK  VIII.  263 

But  OA  =  OB,  therefore  oa  =  oh ;  hence  all  the  straight 
lines  drawn  from  o  to  the  perimeter  of  the  section  are  equal, 
and  the  section  is  a  circle. 

17.  Corollary.  The  axis  of  a  circular  cone  passes  through 
the  centres  of  all  the  sections  parallel  to  the  base, 

18.  Definition.  A  tangent  plane  to  a  cone  is  a  plane  which 
passes  through  an  element  of  the  curved  surface  without 
cutting  this  surface.     The  element  through  which  it  passes 

is  called  the  element  of  contact.  Cv//^^^'^^^ 


THE  SPHERE. 

19.  Definition.  A  sphere  is  a  solid  bounded  by  a  surface  all 
the  points  of  which  are  equally  distant  from  a  point  within, 
called  the  centre. 

A  sphere  may  be  generated  by  the  revolu- 
tion of  a  semicircle  ABC  about  its  diameter 
J.  (7  as  an  axis ;  for  the  surface  generated 
by  the  curve  ABC  will  have  all  its  points 
equally  distant  from  the  centre  0. 

A  radius  of  the   sphere  is  any  straight 
line  drawn  from  the  centre  to  the  surface. 
A  diameter  is  any  straight  line  drawn  through  the  centre  and 
terminated  both  ways  by  the  surface. 

Since  all  the  radii  are  equal  and  every  diameter  is  double 
the  radius,  all  the  diameters  are  equal. 

20.  Definition.  It  will  be  shown  that  every  section  of  a 
sphere  made  by  a  plane  is  a  circle ;  and,  as  the  greatest  pos-- 
sible  section  is  one  made  by  a  plane  passing  through  the 
centre,  such  a  section  is  called  a  great  circle.  Any  section 
made  by  a  plane  which  does  not  pass  through  the  centre  is 
called  a  small  circle. 


264  ELEMENTS   OF   GEOMETRY. 

21.  Definition.  The  poles  of  a  circle  of  the  sphere  are  the 
extremities  of  the  diameter  of  the  sphere  which  is  perpen- 
dicular to  the  plane  of  the  circle  j  and  this  diameter  is  called 
the  axis  of  the  circle. 


PROPOSITION  v.— THEOREM. 

22.  Every  section  of  a  sphere  made  by  a  plane  is  a  circle. 

1st.  If  the -plane  passes    through 
the   centre  of  the   sphere,  the  lines  .^-^^---^ 

joining  points  on  the  perimeter  of  aJifs//.V}liyss^\c 

the  section  with  the  centre  0  of  the        /  ^^^vHI]^?'^  \ 
sphere  are  radii  of  the  sphere,  and      ^F;---— --^------^ 

are  therefore  all  equal.     Consequently        \  j  J 

it  is  a  circle  with  its  centre  at  0.  X,^^   I    ^^/ 

2d.  If   the    plane    does    not    pass  ^ 

through  the  centre  of  the  sphere,  as 

abcj  draw  a  diameter  EOD  pei*pendicular  to  the  section  and 
meeting  it  at  o.  If  points  a,  b,  c,  of  the  perimeter  are  joined 
with  0,  and  also  with  0,  the  triangles  aeO,  boO,  coO,  are  all 
equal  (I.,  Proposition  X.).  Therefore  ao,  bo,  co,  etc.,  are  all 
equal,  and  the  section  is  a  circle  with  its  centre  at  o. 

23.  Corollary  I.  The  axis  of  a  circle  on  a  sphere  passes 
through  the  centre  of  a  .circle. 

24.  Corollary  II.  All  great  circles  of  the  same  sphere  are 
equal. 

25.  Corollary  III.  Every  great  circle  divides  the  sphere  into 
two  equal  parts. 

Suggestion.  Superpose  one  part  upon  the  other,    (v.  II., 
Proposition  II.) 


BOOK  VIII.  265 

26.  Corollary  IY.  Any  two  great  circles  on  the  same  sphere 
bisect  each  other ;  for  the  common  inter- 
section AB  of  their  planes  passes  through 
the  centre  of  the  sphere  and  is  a  diame- 
ter of  each  circle. 

27.  Corollary  Y.  An  arc  of  a  great 
circle  may  he  drawn  through  any  two  given 
points  of  the  surface  of  the  sphere^  and, 
unless  the  points  are  the  opposite  eoctremities 

of  a  diameter,  only  one  such  arc  can  he  drawn;  for  the  two 
points,  together  with  the  centre  0,  determine  the  plane  of  a 
great  circle  whose  circumference  passes  through  the  points. 

If,  however,  the  two  given  points  are  the  extremities  A 
and  5  of  a  diameter  of  the  sphere,  the  position  of  the  circle 
is  not  determined,  for  the  points  J.,  0,  and  B,  being  in  the 
same  straight  line,  will  not  determine  a  plane  (YI.,  Proposi- 
tion I.). 

28.  Corollary  YI.  An  arc  of  a  circle  may  he  drawn  through 
any  three  given  points  on  the  surface  of  the  sphere;  for  the 
three  points  determine  a  plane  which  cuts  the  sphere  in  a 
circle. 

EXEBCISE. 

Theorem. — The  greater  the  distance  of  the  plane  of  a  small 

hsircle  from  the  centre  of  the  sphere,  the  less  the  circle. 

li  23 


266  ELEMENTS  OF   GEOMETRY. 

PROPOSITION  VI.— THEOREM. 

29.  All  the  points  in  the  circumference  of  a  circle  of  the  sphere 
J,-  are  equally  distant  from  either  of  its  poles. 

Let  abed  be  any  circle'of  the  sphere,  ..-s^^^--^ 

and  PP' the  diameter  of  the  sphere  „/^'Wt'^' — ^\j 

perpendicular  to  its  plane;  then,  by  I  %. 
the  definition  (21),  P  and  P'  are  the  ^^^ 
poles  of  the  circle  ahcd^  and,  by  Prop-  \  \^ 
osition  Y.,  Corollary  I.,  FT'  passes  \ 

through  0,  the  centre  of  ahcd.    Join  ^1^ 

P  with  any  points,  a,  6,  c,  on  the  cir- 
cumference of  the  circle.  Then  Pa,  P6,  Pc,  are  equal,  since 
the  triangles  Poa,  Tob^  Poc^  are  equal,  by  I.,  Proposition  VI. 
Hence  all  the  points  of  the  circumference  abed  are  equally 
distant  from  the  pole  P.  For  the  same  reason,  they  are 
equally  distant  from  the  pole  P'. 

30.  Corollary  I.  All  the  arcs  of  great  circles  drawn  from  a 
pole  of  a  circle  to  points  in  its  circumference  are  equal,  since 
their  chords  are  equal  chords  in  equal  circles. 

By  the  distance  of  two  points  on  the  surface  of  a  sphere 
is  usually  understood  the  arc  of  a  great  circle  joining  the  two 
points.  The  arc  of  a  great  circle  drawn  from  any  point  of  a 
given  circle  abc,  to  one  of  its  poles,  as  the  arc  Pa,  is  called 
the  polar  distance  of  the  given  circle,  and  the  distance  from 
the  nearest  pole,  is  usually  understood. 

31.  Corollary  II.  The  polar  distance  of  a  great  circle  is  a 
quadrant ;  thus,  PA,  PB,  etc.,  P'A,  P'B,  etc.,  polar  distances 
of  the  great  circle  ABCD,  are  quadrants ;  for  they  are  the 
measures  of  the  right  angles  A  OP,  BOP,  AOP',  BOP',  etc., 
whose  vertices  are  at  the  centre  of  the  great  circles  PAP'j 
PBP',  etc. 


BOOK  VIII.  267 

In  connection  with  the  sphere,  by  a  quadrant  is  usually  to 
be  understood  a  quadrant  of  a  great  circle. 

32.  Corollary  III.  If  a  point  on  the  surface  of  a  sphere  is 
at  a  quadrant's  distance  from  each  of  two  given  points  of  the 
surface  which  are  not  opposite  extremities  of  a  diameter,  it  is  the 
pole  of  the  great  circle  passing  through  them. 

Suggestion.  Let  P  be  at  a  quadrant's  distance  from  B  and 
C;  then  FOB  and  POC  are  right  angles,  and  FO  is  perpen- 
dicular to  the  plane  ABCD. 

33.  Scholium.  By  means  of  poles,  arcs  of  circles  may  be 
drawn  upon  the  surface  of  a  sphere  with  the  same  ease  as 
upon  a  plane  surface.  Thus,  by  revolving  the  arc  Fa  about 
the  pole  P,  its  extremity  a  will  describe  the  small  circle  abd  ; 
and  by  revolving  the  quadrant  FA  about  the  pole  P,  the 
extremity  A  will  describe  the  great  circle  ABD. 

If  two  points,  B  and  (7,  are  given  on  the  surface,  and  it  is 
required  to  draw  the  arc  ^C,  of  a  great  circle,  between  them, 
it  will  be  necessary  first  to  find  the  pole  P  of  this  circle ;  for 
which  purpose,  take  B  and  G  as  poles,  and  at  a  quadrant's 
distance  describe  two  arcs  on  the  surface  intersecting  in  P. 
The  arc  BC  can  then  be  described  with  a  pair  of  compasses, 
placing  one  foot  of  the  compasses  on  P  and  tracing  the  arc 
with  the  other  foot.  The  opening  of  the  compasses  (distance 
between  their  feet)  must  in  this  case  be  equal  to  the  chord 
of  a  quadrant ;  and  to  obtain  this  it  is  necessary  to  know  the 
radius  of  the  sphere. 

34.  Definition.  A  plane  is  tangent  to  a  sphere  when  it  has 
but  one  point  in  common  with  the  surface  of  the  sphere, 

35.  Definition.  Two  spheres  are  tangent  to  each  other  when 
their  surfaces  have  but  one  point  in  common. 


268 


ELEMENTS  OF   GEOMETRY. 


PROPOSITION  VII.— THEOREM. 

36.  A  plane  tangent  to  a  sphere  is  perpendicular  to  the  radius 
drawn  to  the  point  of  contact. 

For  any  other  line  drawn  from 
the  centre  of  the  sphere  to  the 
plane  must  reach  beyond  the  sur- 
face of  the  sphere,  and  therefore 
must  be  greater  than  the  radius. 
The  radius  is,  then,  the  shortest 
line  that  can  be  drawn  from  the 
centre  of  the  sphere  to  the  plane, 

and  is  consequently  perpendicular  to  the  plane  (YI.,  Proposi- 
tion III.). 

37.  Corollary.   Conversely,   a  plane  perpendicular  to   a 
radius  of  a  sphere  at  its  extremity  is  tangent  to  the  sphere. 

38.  Scholium.  Any  straight  line  A  T,  drawn  in  the  tangent 
plane  through  the  point  of  contact,  is  tangent  to  the  sphere. 

Any  two  straight  lines,  AT,  A T\  tangent  to  the  sphere  at 
the  same  point  A,  determine  the  tangent  plane  at  that  point. 


PROPOSITION  VIII.— THEOREM. 

39.  The  intersection  of  two  spheres  is  a  circle  whose  plane  is 
perpendicular  to  the  straight  line  joining  the  centres  of  the 
spheres,  and  whose  centre  is  in  that  line. 

Through  the  centres  0  and  (7  of 
the  two  spheres  let  any  plane  be 
passed,  cutting  the  spheres  in  great 
circles  which  intersect  each  other  in 
the  points  A  and  B ;  the  chord  AB 
is  bisected  at  G  by  the  line  0(7  at 
right  angles  (II.,  Proposition  YI.,  Corollary  II.).     If  we  now 


Tv^ 


BOOK  VIII.  269 

revolve  the  plane  of  these  two  circles  about  the  line  0(7,  the 
circles  will  generate  the  two  spheres,  and  the  point  A  will 
describe  the  line  of  intersection  of  their  surfaces.  Moreover, 
since  the  line  AC  will,  during  this  revolution,  remain  perpen- 
dicular to  0(7,  it  will  generate  a  circle  whose  plane  is  per- 
pendicular to  0(y,  and  whose  centre  is  C. 

SPHERICAL  ANGLES. 

40.  Definition.  The  angle  of  two  curves  passing  through  the 
same  point  is  the  angle  formed  by  the  two  tangents  to  the 
curves  at  that  point. 

This  definition  is  applicable  to  any  two  intersecting  curves 
in  space,  whether  drawn  in  the  same  plane  or  upon  a  surface 
of  any  kind. 

,  PROPOSITION  IX.— THEOREM. 

M.  The  angle  of  two  arcs  of  great  circles  is  equal  to  the  angle 
of  their  planes^  and  is  measured  by  the  arc  of  a  great  circle 
described  from  its  vertex  as  a  pole  and  included  between  its  sides 
(produced  if  necessary). 

Let  AB  and  AB'  be  two  arcs  of  >^      j, 

great  circles,  A  T  and  A  T'  the  tan-  /         i     \^V~-^ 

gents  to  these  arcs  at  A^  and  0  the         /      j W  \b 

centre  of  the  sphere.    TA  and  T'A      ^L^        4*r~"f3)^ 
lie  in  the  planes  of  their  arcs,  and        \^^~~"^     \       I    / 
are  perpendicular  to  the  radius  OA  X^^    !    J/ 

drawn   to  their  point  of  contact.  ^ 

They  form,  then,  the  plane  angle 

measuring  the  diedral  angle  formed  by  the  planes  of  the  arcs ; 
but,  by  (40),  the  angle  which  they  form  is  equal  to  the  angle 
of  the  two  arcs. 


270 


ELEMENTS   OF  GEOMETRY. 


Now  let  CO'  be  the  arc  of  a  great  circle  described  from  A 
as  a  pole  aiid  intersecting  the  arcs 
AB^  AB'  (produced  if  necessary), 
in  G  and  C.  The  radii  OG  and 
OG'  are  perpendicular  to  J.0,  since 
the  arcs  AG,  AG',  are  quadrants 
(Proposition  YI.,  Corollary  II.); 
therefore  the  angle  GOG'  is  a  plane 
angle  of  the  diedral  angle  A  0,  and 
is  equal  to  TAT',  or  to  BAB',  and  it  is  obviously  measured 
by  the  arc  GG'. 

42.  Corollary.  All  arcs  of  great  circles  drawn  through  the 
pole  of  a  given  great  circle  are  perpendicular  to  its  circumfer- 
ence;  for  their  planes  are  perpendicular  to  its  plane  (YI., 
Proposition  XIY.). 


SPHERICAL  POLYGONS. 

43.  Definition.  A  spherical  polygon  is  a  portion  of  the  sur- 
face of  a  sphere  bounded  by  three  or  more  arcs  of  great 
circles,  as  ABGD. 

Since  the  planes  of  all  great  circles  pass 
through  the  centre  of  the  sphere,  the  planes 
of  the  sides  of  a  spherical  polygon  form,  at 
the  centre  0,  a  polyedral  angle  of  which 
the  edges  are  the  radii  drawn  to  the  ver- 
tices of  the  polygon,  the  face  angles  are 
angles  at  the  centre  measured  by  the  sides  of  the  polygon, 
and  the  diedral  angles  are  equal  to  the  angles  of  the  polygon 
(Proposition  IX.). 

Since  in  a  polyedral  angle  each  face  angle  is  assumed  to  be 
less  than  two  right  angles,  each  side  of  a  spherical  polygon 
will  be  assumed  to  be  less  than  a  semi-circumference. 


BOOK  viir.  271 

A  spherical  polygon  is  convex  when  its  corresponding  poly- 
edral  angle  at  the  centre  is  convex  (VI.,  52). 

A  diagonal  of  a  spherical  polygon  is  an  arc  of  a  great  circle 
joining  any  two  vertices  not  consecutive. 

44.  Definition.  A  spherical  triangle  is  a  spherical  polygon 
of  three  sides.  It  is  called  right  angled^  isosceles^  or  equilat- 
eral^ in  the  same  cases  as  a  plane  triangle. 

45.  In  consequence  of  the  relation  established  between 
polyedral  angles  and  spherical  polygons  (43),  it  follows  that 
from  any  property  of  polyedral  angles  we  may  infer  an  anal- 
ogous property  of  spherical  polygons. 

Eeciprocally,  from  any  property  of  spherical  polygons  we 
may  infer  an  analogous  property  of  polyedral  angles. 

The  latter  is  in  almost  all  cases  the  more  simple  mode  of 
procedure,  inasmuch  as  the  comparison  of  figures  drawn  on 
the  surface  of  a  sphere  is  nearly  if  not  quite  as  simple  as  the 
comparison  of  plane  figures. 

46.  Arcs  of  great  circles  on  the  same  sphere  can  be  super- 
posed and  made  to  coincide  just  as  straight  lines  are  super- 
posed and  made  to  coincide.  We  have  merely  to  place  one 
point  of  the  first  arc  on  some  given  point  of  the  second,  and, 
keeping  this  point  fixed,  to  turn  the  first  arc  about  it  as  a 
pivot,  until  some  second  point  in  the  arc  not  diametrically 
opposite  the  fixed  point  falls  on  the  second  arc.  The  two 
arcs  must  then  coincide  throughout,  by  Proposition  Y.,  Corol- 
lary Y. 

Equal  angles  formed  by  arcs  of  great  circles  on  the  surface 
of  the  same  sphere  can  be  superposed  and  made  to  coincide 
just  as  equal  plane  angles  are  superposed  and  made  to  co- 
incide ;  that  is,  if  the  vertex  of  the  first  angle  is  placed  upon 
the  vertex  of  the  second,  and  one  side  of  the  first  placed 
upon  the  corresponding  side  of  the  second,  the  other  side  of 


272  ELEMENTS   OF   GEOMETRY. 

the  first  will  coincide  with  the  other  side  of  the  second.  For, 
if  the  two  given  angles  are  equal,  their  diedral  angles  are 
equal  (Proposition  IX.).  If  the  vertices  of  the  angles  co- 
incide, the  edges  of  the  diedral  angles  coincide ;  if  a  side  of 
the  first  angle  is  placed  on  a  side  of  the  second,  one  face  of 
the  first  diedral  angle  coincides  with  one  face  of  the  second. 
The  remaining  faces  of  the  diedral  angles  must  then  coincide, 
and  consequently  the  remaining  sides  of  the  given  angles 
coincide. 

47.  Definition.  Two  spherical  triangles  are  symmetrical  if 
all  the  parts  of  one  are  respectively  equal  to  the  parts  of  the 
other,  but  the  corresponding  parts  are  arranged  in  opposite 
orders  in  the  two  triangles. 

Two  symmetrical  triangles,  as  ABC^  ABC'j 
in  the  figure  cannot  be  made  to  coincide; 
for,  to  bring  the  vertex  C  upon  the  corre- 
sponding vertex  C,  the  second  triangle  would 
have  to  be  turned  over,  and  the  two  convex 
surfaces  would  thus  be  brought  together. 

48;^here  is,  however,  one  exception  to  the 
last  remark, — namely,  the  case  of  symmetrical  isosceles  trian- 
gles, Por,  if  ABC  is  an  isosceles 
spherical  triangle  and  AB  =  AC, 
then,  in  its  symmetrical  triangle, 
we  have  A'B'  =  A!C\  and  con- 
sequently AB  =  A'C,  AG  = 
A'B\  and,  since  the  angles  A  and 

A'  are  equal,  if  AB  be  placed  on  A'G\  AG  will  fall  on  its 
equal  A'B\  and  the  two  triangles  will  coincide  through- 
out. 

49.  Definition.  If  from  the  vertices  of  a  spherical  triangle 
as  poles,  arcs  of  great  circles  are  described,  these  arcs  form 


/}l\    / 


4 


BOOK   VIII. 


273 


by  their  intersection  a  second  triangle,  which  is  called  the 
polar  triangle  of  the  first. 

Thus,  if  Aj  B,  and  C  are  the  poles  of  the 
arcs  of  great  circles,  B'C'j  A'C\  and  A'B\ 
respectively,  A'B'C  is  the  polar  triangle  of 
ABC. 

Since   all  great   circles,   when  completed, 
intersect  each  other  in  two  points,  the  arcs 
jB'C",  A'C,  A'B\  if  produced,  will  form  three 
other  triangles ;  but  the  triangle  which  is  taken  as  the  polar 
triangle  is  that  whose  vertex  A\  homologous  to  A,  lies  on 
the  same  side  of  the  arc  BC  as  the  vertex  A;  and  so  of  the 
other  vertices. 


PROPOSITION  X.— THEOREM. 

50y^f  the  first  of  two  spherical  triangles  is  the  polar  triangle 
of  the  second,  then,  reciprocally,  the  second  is  the  polar  triangle 
of  the  first. 

Let  A'B'C  be  the  polar  triangle  of  ABC; 
then  is  ABC  the  polar  triangle  of  A'B'C, 

For,  since  A  is  the  pole  of  the  arc  B'C, 
the  point  5'  is  at  a  quadrant's  distance  from 
A ;  and,  since  C  is  the  pole  of  the  arc  A'B', 
the  point  ^'  is  at  a  quadrant's  distance  from 
C ;  therefore  B'  is  the  pole  of  the  arc  AC 
(Proposition  YI.,  Corollary  III.).  In  the  same  manner  it  is 
shown  that  A'  is  the  pole  of  the  arc  BC,  and  C"  the  pole  of 
the  arc  AB.  Moreover,  A  and  A'  are  on  the  same  side  of 
B'C\  B  and  B'  on  the  same  side  of  A'C,  C  and  C  on  the 
same  side  of  A'B' ;  therefore  ABC  is  the  polar  triangle  of 
A'B'C. 


274 


ELEMENTS   OF   GEOMETRY. 


PROPOSITION  XI.— THEOREM. 

5lX/«  t.o  polar  triangles,  ea^U  anyle  of  one  is  measure,  l>y 
the  supplement  of  the  side  lying  opposite  to  it  in  the  other. 

Let  ABC  and  A'B'C  be  two  polar  tri- 
angles. 

Let  the  sides  AB  and  AC,  produced  if 
necessary,  meet  the  side  B'C  in  the  points  h 
and  c.  The  vertex  A  being  the  pole  of  the 
arc  he,  the  angle  A  is  measured  by  the  arc  he 
(Proposition  IX.). 

Now,  B '  being  the  pole  of  the  arc  Ac,  and 
C  the  pole  of  the  arc  Ah^  the  arcs  B'c  and  G'h  are  quadrants ; 
hence  we  have 

B'C  -\-  he  =^B'c  -\-  C'h  =  Si  semi-circumference. 

Therefore  he,  which  measures  the  angle  A,  is  the  supplement 
of  the  side  B'G\ 

In  the  same  manner  it  can  be  shown  that  each  angle  of 
either  triangle  is  measured  by  the  supplement  of  the  side 
lying  opposite  to  it  in  the  other  triangle. 

.52.  Scholium.  Let  the  angles  of  the  trian- 
gle ABC  be  denoted  by  A,  B,  and  C,  and  let 
the  sides  opposite  to  them,  namely,  BC,  AC, 
and  AB,  be  denoted  by  a,  h,  and  c,  respec- 
tively. Let  the  corresponding  angles  and 
sides  of  the  polar  triangle  be  denoted  by  A', 
B',  C,  a',  h',  and  c\  Also  let  both  angles  and 
sides  be  expressed  in  degrees.  Then  the  preceding  theorem 
gives  the  following  relations : 

A  +a'=B  -\-h':=^C  +c'^  180°. 
A'  +a  :=:^B'  -\rh  =C'  +  c  =  180°. 


BOOK  VIII.  275 


PROPOSITION  XII.— THEOREM. 

53.  Two  triangles  on  the  same  sphere  are  either  equal  or  sym- 
metrical, when  two  sides  and  the  included  angle  of  one  are  re- 
spectively equal  to  txoo  sides  and  the  included  angle  of  the  other. 

In  the  triangles  ^50  and  DEF,  let  the 
angle  A  be  equal  to  the  angle  D,  the  side 
AB  equal  to  the  side  DE,  and  the  side 
A  G  equal  to  the  side  DF. 

1st.  When  the  parts  of  the  two  trian- 
gles are  in  the  same  order,  ABC  can  be 
applied  to  DEF^  as  in  the  corresponding 
case  of  plane  triangles  (I.,  Proposition 
VI.),  and  the  two  triangles  will  coincide ;  therefore  they  are 
equal. 

2d.  When  the  parts  of  the  two 
triangles  are  in  inverse  order,  let 
DE'F  be  the  symmetrical  triangle 
of  BEF,  and  therefore  having  its 
angles  and  sides  equal,  respectively, 
to  those  of  DEF.  Then,  in  the 
triangles  ABC  and  DE'F,  we  shall 
have  the  angle  BAC  equal  to  the 

angle  E'DF,  the  side  AB  to  the  side  DE',  and  the  side 
AG  to  the  side  DF,  and  these  parts  arranged  in  the  same 
order  in  the  two  triangles ;  therefore  the  triangle  ABG  is 
equal  to  the  triangle  DE'F,  and  consequently  symmetrical 
with  DEF, 

54.  Scholium.  In  this  proposition,  and  in  the  propositions 
which  follow,  the  two  triangles  may  be  supposed  on  the  same 
sphere,  or  on  two  equal  spheres. 


276 


ELEMENTS   OF  GEOMETRY. 


PKOPOSITION  XIII.— THEOREM. 

55.  Two  triangles  on  the  same  sphere  are  either  equal  or  sym- 
metrical^  when  a  side  and  two  adjacent  angles  of  one  are  equal 
respectively  to  a  side  and  two  adjacent  angles  of  the  other. 

Let  the  triangles  ABC 
and  BEF  have  the  side  a 
equal  to  the  side  d^  and  the  c'  /  "^  X6' 
angles  B^  C,  equal  respec- 
tively to  the  angles  E,  F ; 
then  are  the  triangles  equal. 

Construct  the  polar  tri- 
angles of  ABC  and  DEF.    We  have  h'  =  e',  c'  =  /',  and  A'  == 
D\  by  Proposition  XI.     Then  A'B'C  and  D'E'F'  are  equal 
or  symmetrical,  by  Proposition  XII.     Therefore  their  polar 
triangles  ABC,  DEF,  are  equal  or  symmetrical. 

58.  Scholium.  The  proposition  might  be  proved  by  direct 
superposition,  as  in  I.,  Proposition  YII. 


PROPOSITION  XIV.— THEOREM. 

57.  Two  triangles  on  the  same  sphere  are  either  equal  or  sym- 
metrical, when  the  three  sides  of  one  are  respectively  equal  to  the 
three  sides  of  the  other. 

For  if  their  vertices  are  joined  with  the  centre  of  the 
sphere,  the  triedral  angles  thus  formed  have  the  three  face 
angles  of  the  one  respectively  equal  to  the  three  face  angles 
of  the  other,  and  consequently,  by  VI.,  Proposition  XXII.,  their 
corresponding  diedral  angles  are  equal.  The  given  triangles 
are,  then,  mutually  equilateral  and  mutually  equiangular,  and 
are  equal  or  symmetrical. 


i 


BOOK  vni.  277 

58.  Scholium.  The  proposition  can  be  proved  as  in  I.,  Prop- 
osition IX. 


PROPOSITION  XV.— THEOREM. 

59.  If  two  triangles  on  the  same  sphere  are  mutually  equi- 
angular, they  are  also  mutually  equilateral,  and  are  either  equal 
or  symmetrical.  . 

Let  the  spherical  trian-  '^'"'^  ^^^^^ 

gles  M  and  JV  be  mutually 
equiangular. 

Let  M'  be  the  polar  tri- 
angle of  M,  and  iV'  the  polar  triangle  of  iV.  Since  M  and 
N  are  mutually  equiangular,  their  polar  triangles  M'  and  iV' 
are  mutually  equilateral  (Proposition  XL);  therefore,  by 
Proposition  XIY.,  the  triangles  M^  and  N'  are  mutually  equi- 
angular. But  M'  and  iV'  being  mutually  equiangular,  their 
polar  triangles  M  and  iV  are  mutually  equilateral.  Conse- 
quently, M  and  N  are  either  equal  or  symmetrical. 

60.  Scholium.  It  may  seem  to  the  student  that  the  pre- 
ceding property  destroys  the  analogy  which  subsists  be^^een 
plane  and  spherical  triangles,  since  two  mutually  equiangular 
plane  triangles  are  not  necessarily  mutually  equilateral.  But 
in  the  case  of  spherical  triangles  the  equality  of  the  sides 
follows  from  that  of  the  angles  only  upon  the  condition  that 
the  triangles  are  constructed  upon  the  same  sphere  or  on 
equal  spheres ;  if  they  are  constructed  on  spheres  of  diiferent 
radii,  the  homologous  sides  of  two  mutually  equiangular  tri- 
angles will  no  longer  be  equal,  but  will  be  proportional  to  the 
radii  of  the  sphere ;  the  two  triangles  will  then  be  similar,  as 
in  thje  case  of  plane  triangles. 

24 


278  ELEMENTS   OF   GEOMETRY. 

EXERCISES. 

1.  Theorem. — In  an  isosceles  spherical  triangle  the  angles 
opposite  the  equal  sides  are  equal. 

2.  Theorem. — The  arc  drawn  from  the  vertex  of  an  isosceles 
spherical  triayigle  to  the  middle  point  of  the  base  is  perpendicular 
to  the  base,  and  bisects  the  vertical  angle. 

3.  Theorem. — If  two  angles  of  a  spherical  triangle  are  equal, 
the  triangle  is  isosceles. 

PROPOSITION  XVI.-^THEOREM. 

61.  Any  side  of  a  spherical  triangle  is  less  than  the  sum  of 

the  other  two. 

Let  ABC  be  a  spherical  triangle; 
then  any.  side,  as  AC,  is  less  than  the 
sum  of  the  other  two,  AB  and  BC.  / 

For,   in   the    corresponding   triedral  / 

angle  formed  at  the  centre  0  of  the  /^'l-"'''' 

sphere,  we  have  the  angle  AOC  less      o**' 
than  the  sum  of  the  angles  AOB  and 

BOC  (VI.,  Proposition  XX.)  ;  and  since  the  sides  of  the  tri- 
angle measure  these  angles,  respectively,  we  have  AC  <^  AB 
+  BC. 

EXERCISES. 

1.  Theorem. — If  two  angles  of  a  spherical  triangle  are  un- 
equal, the  side  opposite  the  greater  angle  is  greater  than  the  side 
opposite  the  less  angle,     (v.  I.,  Proposition  XII.) 

2.  Theorem. — If  two  sides  of  a  spherical  triangle  are  unequal, 
the  angle  opposite  the  greater  side  is  greater  than  the  angle  oppo- 
site the  less  side.     (v.  L,  Proposition  XIII.) 


BOOK   VIIT. 


279 


PEOPOSITION  XVII.— THEOEEM. 

62.  The  sum  of  the  sides  of  a  convex  spherical  polygon  is  less 
than  the  circumference  of  a  great  circle. 

For  the  sum  of  the  face  angles  of  the  corresponding  poly- 
edral  angle  at  the  centre  of  the  sphere  is  less  than  four  right 
angles  (YI.,  Proposition  XXI.). 


PEOPOSITION  XVIII.— THEOEEM. 

^  63.  The  sum  of  the  angles  of  a  spherical  triangle  is  greater 
than  two,  and  less  than  six,  right  angles. 

For,  denoting  the  angles  of  a  spherical  tri- 
angle by  A,  B,  C,  and  the  sides  respectively 
opposite  to  them  in  its  polar  triangle  by  a',      ^< 
h',  d,  we  have  (Proposition  XI.) 

A  =  180°  —  o!,  J?=:  180°  —  6',  0=  180°  —c', 


the  sum  of  which  is 


A^  B  -^C  ^  540°  —  {a!  +  6'  +  c'). 

But  a'  ^y  -\-  c'  <i  360°  (Proposition  XYII.)  ;  therefore  A  + 
B  -\-  G  ^  180° ;  that  is,  the  sum  of  the  three  angles  is 
greater  than  two  right  angles.  Also,  since  each  angle  is 
less  than  two  right  angles,  their  sum  is  less  than  six  right 
angles. 

64.  Scholium.  A  spherical  triangle  may  have  two  or  even 
three  right  angles  j  also  two  or  even  three  obtuse  angles. 


280 


ELEMENTS  OF  GEOMETRY. 


PROPOSITION  XIX.-THEOEEM. 

65.  Two  symmetrical  spherical  triangles  are  equivalent. 

Let  ABG  and  A'B'C  be  symmetrical  spherical  triangles. 
Let  P  be  the  pole  of  the 
small  circle  passing  through 
A,  B,  and  C  (Proposition 
Y.,  Corollary  YI.).  Then 
the  arcs  PA,  PB,  PC,  are 
equal  (Proposition  YL,  Cor- 
ollary I.),  and  divide  ABG 
into  three  isosceles  trian- 
gles. 

Through  A'  and  B'  in  the  triangle  A'B'C  draw  arcs  mak- 
ing with  A'B'  angles  equal  respectively  to  PAB  and  PBA, 
and  join  P\  their  point  of  intersection,  with  G'.  The  isos- 
celes triangle  PAB  is  equal  to  the  triangle  P'A'B\  by  Propo- 
sition XIII.  and  (48).  The  isosceles  triangle  PBG  is  equal 
to  the  triangle  P'B'G\  by  Proposition  XII.  and  (48).  The 
isosceles  triangle  PGA  is  equal  to  the  triangle  P'G'A\  by 
Proposition  XIY.  and  (48).  Hence  ABG  and  A'B'C'  are 
equivalent. 

If  the  pole  P  should  fall 
without  the  triangle  ABG,  the 
triangle  would  be  equivalent 
to  the  sum  of  two  of  the  isos- 
celes triangles  diminished  by 
the  third ;  but,  as  the  same 
thing  would  occur  for  the  sym- 
metrical triangle,  the  conclu- 
sion would  be  the  same. 


■\p 


pv— -/-- 


BOOK   VIII. 


281 


66.  Definition.  If  a  spherical  triangle  ABC 
has  two  right  angles,  B  and  (7,  it  is  called  a 
hi-rectangular  triangle ;  and,  since  the  sides  AB 
and  J. (7  must  each  pass  through  the  pole  of  BO 
(Proposition  IX.,  Corollary),  the  vertex  A  is 
that  pole,  and  therefore  AB  and  A  C  are  quadrants. 

67.  Definition.  A  lune  is  a  portion  of  the  sur- 
face of  a  sphere  included  between  two  semi- 
circumferences  of  great  circles ;  as  AMBNA. 

The  two  angles  of  a  lune  are  equal,  since 
each  is  equal  to  the  diedral  angle  formed  by 
the  planes  of  the  arcs  of  the  lune ;  and  the 
lune  is  equal  to  the  sum  of  two  equal  bi-rectan- 
gular  triangles,  each  of  which  has  the  angle  of  the  lune  for  its 
third  angle. 

EXERCISE. 

Theorem. — Two  lunes  on  the  same  sphere  or  on  equal  spheres 
re  equal  if  their  angles  are  equal. 

PROPOSITION  XX.— THEOREM.  ^ 

68.  If  two  arcs  of  great  circles  intersect  on  the  surface  of  a 
hemisphere^  the  sum  of  the  opposite  spherical  triangles  which  they 
form  is  equivalent  to  a  lune  whose  angle  is  the  angle  between  the 
arcs  in  question. 

Let  the  arcs  ACA',  BCB\  intersect 
on  the  surface  of  the  hemisphere 
ABA'B'C.  Then  will  the  triangles 
ABG^  AIB'C^  be  together  equivalent 
to  a  lune  whose  angle  is  A  GB. 

For,  continue  the  arcs  ACA\  BCB\ 
until  they  intersect  in  C  A' C  =  AC, 
B'C  ^^  BCy  and  A'B'  =  AB,  since  they  subtend  equal  angles. 


282 


ELEMENTS  OF  GEOMETRY. 


The  triangles  A'B'C  and  ABC  are 
then  equal  or  symmetrical,  by  Propo- 
sition XIY.,  and  are  in  either  case 
equivalent  (Proposition  XIX.).  There- 
fore ABC  and  A'B'C  are  together 
equivalent  to  A'B'C  +  A'B'C;  that 
is,  to  the  lune  CA'C'B\ 


MEASUEEMENT  OF   THE   SUEFACES  OF  SPHEEICAL 

FIGUEES. 

69.  Definition.  A  degree  of  spherical  surface,  or,  more  briefly, 
u  spherical  degree,  is  -^  of  the  surface  of  a  hemisphere.  It 
is  a  convenient  unit  in  measuring  the  surfaces  of  spherical 
figures.  Like  the  degree  of  arcs,  it  is  not  a  unit  of  absolute 
magnitude,  but  depends  upon  the  size  of  the  sphere  on  which 
the  figures  are  drawn. 

It  may  be  conveniently  conceived  as  a  bi-rectangular  spher- 
ical triangle  whose  third  angle  is  an  angle  of  one  degree. 


PEOPOSITION  XXI.— THEOEEM. 

70.  A  lune  is  to  the  surface  of  the  sphere  as  the  angle  of  the 
lune  is  to  four  right  angles. 

Let  AWBMA  be  a  lune,  and  let 
MNP  be  the  great  circle  whose  poles 
are  the  extremities  of  the  diameter 
AB. 

Since  the  angle  of  the  lune  is  meas- 
ured by  the  arc  MW,  the  angle  of  the 
lune  is  to  four  right  angles  as  the  arc 
MN  is  to  the  whole  circumference  MNPM. 

Ist.  Suppose  that  MN  and  the  circumference  have  a  com- 


BOOK  VIII.  283 

mon  measure  which  is  contained  m  times  in  MN  and  n  times 
in  MNPM.     Then 

MN     _  m 
MNPM~~  n 

Apply  the  measure  to  the  circumference,  and  through  the 
points  of  division  and  the  axis  AB  pass  planes;  they  will 
divide  the  whole  surface  of  the  sphere  into  n  equal  lunes  (67, 
Exercise),  of  which  the  given  lune  ANBMA  will  contain  m. 

Therefore 

ANBMA         _m 
surface  of  sphere       w' 
and  we  have 

ANBMA  MN- 


surface  of  sphere       MNFM' 

2d.  We  can  extend  the  proof  to  the  case  where  MN  and 
MNPM  are  incommensurable  by  our  usual  method,  (v.  YII., 
Proposition  YII.) 

71.  Corollary.  The  area  of  a  lune  is  expressed  by  twice  its 
angle,  the  angular  unit  being  the  degree,  and  the  unit  of  surface 
the  spherical  degree. 

For,  by  (69),  the  area  of  the  surface  of  the  sphere  is  720 
spherical  degrees.     We  have,  then,  if  S  is  the  area  and  A 

the  angle  of  the  lune, 

S  ^   A   , 
720       360 ' 
whence 

S=2A. 

72.  Scholium.  If  the  angle  A  contains  a  whole  number  of 
degrees,  and  each  of  the  parts  of  the  arc  MN  in  the  figure 
above  is  one  degree,  each  of  the  small  lunes  is  made  up  of 
two  spherical  degrees,  and  the  lune  AMBN  obviously  con- 
tains twice  as  many  spherical  degrees  as  the  arc  MN  contains 
degrees  of  arc. 


284  ELEMENTS   OF   GEOMETRY. 

A-    PROPOSITION  XXII.— THEOREM. 

73.  The  area  of  a  spherical  triangle  is  equal  to  the  excess  of 
the  sum  of  its  angles  over  two  right  angles. 

For,  let  ABC  be  a  spherical  triangle. 
Complete  the  great  circle  ABA^B',  and 
produce  the  arcs  AC  and  BC  to  meet 
this  circle  in  A'  and  B\ 

We  have,  by  the  figure, 

ABC  +  A'BC  =  lune  A, 
ABC  +  AB'C=luneB, 

and,  by  Proposition  XX., 

ABC  +  A'B'C  =  lune  C. 

The  sum  of  the  first  members  of  these  equations  is  equal  to 
twice  the  triangle  ABC,  plus  the  four  triangles  ABC,  A'BC, 
AB'C,  A'B'C,  which  compose  the  surface  of  the  hemisphere, 
whose  area  is  360  spherical  degrees. 

Therefore,  denoting  the  area  of  the  triangle  ABC  by  T, 
we  have  (Proposition  XXI.,  Corollary) 

2T  +  360°  =  2A-\-2B  +  2(7, 
T  +  180°  =  A  +  B-^  C, 
T  =  A-i-  B  +  C—  180°. 

74.  Scholium.  The  excess  of  the  sum  of  the  angles  of  a 
spherical  triangle  over  two  right  angles  is  sometimes  called 
its  spherical  excess. 

EXERCISE. 

Theorem. — The  area  of  a  spherical  polygon 
is  measured  by  the  sum  of  its  angles  minus 
the  product  of  two  right  angles  multiplied  by 
the  number  of  sides  of  the  polygon  less  two. 


BOOK  VIII.  285 

75.  Scholium.  It  must  not  be  forgotten  that  Propositions 
XXI.  and  XXII.  merely  enable  us  to  express  our  areas  in 
spherical  degrees ;  that  is,  in  terms  of  yj^  of  the  surface  of 
the  whole  sphere.  If  the  area  is  required  in  terms  of  the 
ordinary  unit  of  surface  (lY.,  1),  the  area  of  the  surface  of 
the  sphere  must  first  be  given  in  terms  of  the  unit  in  question. 


SHORTEST    LINE    ON    THE    SURFACE    OF  A  SPHERE 
BETWEEN  TWO   POINTS. 

PROPOSITION  XXIII.— THEOREM. 

76.  The  shortest  line  that  can  be  drawn  on  the  surface  of  a 
sphere  between  two  points  is  the  arc  of  a  great  circle^  not  greater 
than  a  semi-circumference,  joining  the  two  points. 

Let  AB  be  an  arc  of  a  great  circle,  less 
than  a  semi-circumference,  joining  any 
two  points  A  and  B  of  the  surface  of  a 
sphere  ;  and  let  C  be  any  arbitrary  point 
taken  in  that  arc.  Then  we  say  that 
the  shortest  line  from  A  to  B,  on  the  sur- 
face of  the  sphere,  must  pass  through  C. 

From  A  and  B  as  poles,  with  the  polar  distances  A  C  and 
BCj  describe  circumferences  on  the  surface  ',  these  circumfer- 
ences touch  at  C  and  lie  wholly  without  each  other.  For, 
let  Mhe  any  point  other  than  C  in  the  circumference  whose 
pole  is  A,  and  draw  the  arcs  of  great  circles  AM,  BM,  form- 
ing the  spherical  triangle  A  MB.  We  have,  by  Proposition 
XYI.,  AM  +  BM  >  AB,  and  subtracting  from  the  two 
members  of  this  mequality  the  equal  arcs  AM  and  AC,  wq 
have  BM  >  BC;  therefore  ikf  lies  without  the  circumference 
whose  pole  is  B. 


^ 


286  ELEMENTS  OF  GEOMETRY. 

Now  let  AFGB  be  any  line  from  A  to  5,  on  the  surface  of 
the  sphere,  which  does  not  pass  through  the  point  C,  and 
which  therefore  cuts  the  two  circumfer- 
ences in  different  points,  one  in  F^  the 
other  in  G.  Then  a  shorter  line  can  be 
drawn  from  A  to  B^  passing  through  G. 
For,  whatever  may  be  the  nature  of  the 
line  AF^  an  equal  line  can  be  drawn  from 
Ato  C;  since,  if  AC  and  AF  be  conceived 
to  be  drawn  on  two  equal  spheres  having 
a  common  diameter  passing  through  A,  and  therefor"  having 
their  surfaces  in  coincidence,  and  if  one  of  these  spfh^res  be 
turned  upon  the  common  diameter  as  an  axis,  the  point  A 
will  be  fixed  and  the  point  F  will  come  into  coincidence  with 
C;  the  surfaces  of  the  two  spheres  continuing  to  coincide, 
the  line  AF  will  then  lie  on  the  common  surface  between  A 
and  C.  For  the  same  reason,  a  line  can  be  drawn  from  B  to 
0,  equal  to  BG.  Therefore  a  line  can  be  drawn  from  A  to  B, 
through  G,  equal  to  the  sum  of  AF  and  BG,  and  consequently 
less  than  AFGB.  The  shortest  line  from  A  to  B  therefore 
passes  through  C;  that  is,  through  any,  or  every,  point  in 
AB.     Consequently  it  must  be  the  arc  AB  itself. 


EXERCISES  ON  BOOK  VIII. 


THEOREMS. 


1.  A  SPHERE  can  be  circumscribed  about  any  tetraedron. 
Suggestion.    The    locus  of    the    points 

equally  distant  from  A,  B^  and  C  is  the 
perpendicular  EM  erected  at  the  centre 
of  the  circle  circumscribed  about  ABC 
(VI.,  Exercise  15.)  The  locus  of  the 
points  equally  distant  from  B^  C,  and  D 
is  tne  perpendicular  FN^  and  both  EM 
and  FN  lie  in  the  plane  perpendicular  to 
BC  Sit  its  middle  point,  since  that  plane 
contains  all  the  points  equally  distant 
from  B  and  C.  EM  and  FN  therefore 
intersect,  and  O,  their  point  of  intersec- 
tion, is  equally  distant  from  the  four  ver- 
tices of  the  tetraedron.  There  is  only  one  such  point.  Therefore 
only  one  sphere  can  be  circumscribed  about  a  tetraedron. 

2.  The  perpendiculars  erected  at  the  centres  of  the  four  faces 
of  a  tetraedron  meet  in  a  point. 


3.  A  sphere  can  be  inscribed  in  any  tetra- 
edron. 

Suggestion,  The  locus  of  the  points  equally 
distant  from  two  faces  of  the  tetraedron  is 
the  plane  bisecting  the  diedral  angle  be- 
tween them. 


4.  The  planes  bisecting  the  six  diedral  angles  of  a  tetraedron 
intersect  in  a  point. 

287 


288  ELEMENTS   OF   GEOMETRY. 


LOCI. 

6.  Locus  of  the  points  in  space  which  are  at  a  given  distance 
from  a  given  straight  line. 

6.  Locus  of  the  points  which  are  at  the  distance  a  from  a  point 
Ay  and  at  the  distance  b  from  a  point  B. 

7.  Locus  of  the  centres  of  the  spheres  which  are  tangent  to 
three  given  planes. 

8.  Locus  of  the  centres  of  the  sections  of  a  given  sphere  made 
by  planes  passing  through  a  given  straight  line. 

Suggestion.  Pass  a  plane  through  the  centre  of  the  sphere  per- 
pendicular to  the  given  straight  line.    Then  see  II.,  Exercise  24. 

9.  Locus  of  the  centres  of  the  sections  of  a  given  sphere  made 
by  planes  passing  through  a  given  point. 


PROBLEMS. 


10.  Through  a  given  point  on  the  surface  of  a  sphere,  to  pass  a 
plane  tangent  to  the  sphere,    (v.  Proposition  VII.,  Corollary.) 

11.  Through  a  given  straight  line  without  a  sphere,  to  pass  a 
plane  tangent  to  the  sphere. 

Suggestion.  Through  the  centre  of  the  sphere  pass  a  plane  per- 
pendicular to  the  given  line.  In  this  plane,  from  its  point  of 
intersection  with  the  line,  draw  a  line  tangent  to  the  circle  in 
which  the  plane  cuts  the  sphere.  A  plane  through  the  tangent 
line  and  the  given  line  is  the  tangent  plane  required.  (Two  solu- 
tions.) 

12.  Through  a  given  point  without  a  sphere,  to  pass  a  plane 
tangent  to  the  sphere. 

13.  To  cut  a  given  sphere  by  a  plane  passing  through  a  given 
straight  line  so  that  the  section  shall  have  a  given  radius. 

SuggeMion.  Pass  a  plane  through  the  centre  of  the  sphere  per- 
pendicular to  the  given  line.    Then  v.  II.,  Exercise  37. 


BOOK  VIII.  289 

14.  To  construct  a  spherical  surface  with  a  given  radius— Ist, 
passing  through  three  given  points ;  2d,  passing  through  two 
given  points  and  tangent  to  a  given  plane,  or  to  a  given  sphere  ; 
3d,  passing  through  a  given  point  and  tangent  to  two  given  planes, 
or  to  two  given  spheres,  or  to  a  given  plane  and  a  given  sphere ; 
4th,  tangent  to  three  given  planes,  or  to  three  given  spheres,  or  to 
two  given  planes  and  a  given  sphere,  or  to  a  given  plane  and 
two  given  spheres. 

15.  Through  a  given  point  on  the  surface  of  a  sphere,  to  draw  a 
great  circle  tangent  to  a  given  small  circle. 

Suggestion,  With  the  pole  of  the  small  circle  as  a  pole,  and  with 
a  polar  distance  equal  to  the  polar  distance  of  the  small  circle 
plus  a  quadrant,  describe  a  second  small  circle.  With  the  given 
point  as  a  pole  describe  a  great  circle.  A  point  of  intersection  of 
this  great  circle  with  the  second  small  circle  will  be  the  pole  of 
the  great  circle  required. 

16.  To  draw  a  great  circle  tangent  to  two  given  small  circles. 

17.  At  a  given  point  in  a  great  circle,  to  driaw  an  arc  of  a  great 
circle  which  shall  make  a  given  angle  with  the  first. 


b 


BOOK  IX. 


MEASUREMENT  OP  THE  THREE  ROUND  BODIES. 
THE  CYLINDER. 

1.  Definition.  The  area  of  the  convex,  or  lateral,  surface 
of  a  cylinder  is  called  its  lateral  area. 

2.  Definition.  A  prism  is  inscribed 
in  a  cylinder  when  its  base  is  inscribed 
in  the  base  of  the  cylinder  and  its 
lateral  edges  are  elements  of  the  cyl- 
inder. It  follows  that  the  upper  base 
of  the  prism  is  inscribed  in  the  upper 
base  of  the  cylinder. 

To  inscribe,  then,  a  prism  of  any 
given  number  of  lateral  faces  in  a  cylinder,  we  have  merely  to 
inscribe  in  the  base  a  polygon  of  the  given  number  of  sides, 
and  through  the  vertices  of  the  polygon  to  draw  elements 
of  the  cylinder.  Planes  passed  through  adjacent  elements 
will  form  the  lateral  faces  of  the  prism  which  is  obviously 
wholly  contained  in  the  cylinder. 

3.  Definition.  A  prism  is  circum- 
scribed about  a  cylinder  when  its 
base  is  circumscribed  about  the 
base  of  the  cylinder  and  its  lateral 
edges  are  parallel  to  elements  of 
the  cylinder. 

It  follows  that  its  lateral  faces 
are  tangent  to  the  lateral  faces  of 
the  cylinder  (YIII.,  11) ;   for  any 

face,  as  AB'y  contains  the  element  bb\  since  it  contains  the 
290 


BOOK   IX. 


291 


parallel  line  AA^  and  the  point  b  (YI.,  Proposition  II.),  and, 
by  YIII.,  Proposition  I.,  it  cannot  cut  the  surface  of  the  cyl- 
inder again  jinless  AB  cuts  the  base  again ;  and  that  its  upper 
base  is  circumscribed  about  the  upper  base  of  the  cylinder. 
The  cylinder  is  obviously  wholly  contained  in  the  prism. 


4.  Definition.  A  right  section  of  a 
cylinder  is  a  section  made  by  a  plane 
perpendicular  to  its  elements ;  asabcdef. 

The  intersection  of  the  same  plane 
with  an  inscribed  or  circumscribed 
prism  is  a  right  section  of  the  prism. 


5.  Definition.  Similar  cylinders  of  revolution  are  those  which 
are  generated  by  similar  rectangles  revolving  about  homolo- 
gous sides. 

PKOPOSITION  I.— THEOREM. 

6.  If  a  prism  whose  base  is  a  regular  polygon  be  inscribed 
in  or  circumscribed  about  a  given  cylinder^  its  volume  will  ap- 
proach the  volume  of  the  cylinder  as  its  limit,  and  its  lateral  sur- 
face will  approach  the  lateral  surface  of  the  cylinder  as  its  limit 
as  the  number  of  sides  of  its  base  is  indefinitely  increased. 

For,  if  we  could  make  the  base 
of  the  prism  exactly  coincide  with 
the  base  of  the  cylinder,  the  prism 
and  the  cylinder  would  coincide 
throughout,  and  their  volumes 
would  be  equal  and  their  lateral 
surfaces  equal. 

But,  by  increasing  the  number 
of  sides  of  the  base  of  the  prism. 


292 


ELEMENTS   OF  GEOMETRY. 


we  can  make  it  come  as  near  as  we  please  to  coinciding  with 
the  base  of  the  cylinder  (Y.,  Propo- 
sition YII.) ;  we  can  then  make 
the  prism  and  cylinder  fail  of  co- 
incidence by  as  small  an  amount 
as  we  choose.  Consequently,  by 
increasing  at  pleasure  the  number 
of  sides  of  the  base  of  the  circum- 
scribed or  inscribed  prism,  we  can 
make  the  difference  between  the 
volumes  of  prism  and  cylinder, 
and  between   the  lateral  surfaces 

of  prism  and  cylinder,  as  small  as  we  choose,  but  cannot  make 
it  absolutely  zero. 

7.  Scholium.  The  proposition  just  proved  is  true  when  the 
base  of  the  prism  is  not  a  regular  polygon ;  but  it  is  only  for 
the  case  of  the  regular  polygon  that  a  rigorous  proof  has 
been  given  in  Book  V. 

PEOPOSITION  II.— THEOREM. 

8.  The  lateral  area  of  a  cylinder  is  equal  to  the  product  of 
the  perimeter  of  a  right  section  of  the  cylinder  by  an  element  of 
the  surface. 

Let  ABCBEF  be  the  base  and  AA' 
any  element  of  a  cylinder,  and  let  the 
curve  abcdef  be  any  right  section  of 
the  surface.  Denote  the  perimeter  of 
the  right  section  by  P,  the  element 
AA'  by  E^  and  the  lateral  area  of  the 
cylinder  by  S. 

Inscribe  in  the  cylinder  a  prism 
ABCDEF  A'  of  any  arbitrarily  chosen 


BOOK  rx. 


293 


number  n  of  faces.  The  right  section,  abcdef,  of  this  prism 
will  be  a  polygon  inscribed  in  the  right  section  of  the  cylin- 
der formed  by  the  same  plane  (4).  Denote  the  lateral  area 
of  the  prism  by  s,  and  the  perimeter  of  its  right  section  by 
p ;  then,  the  lateral  edge  of  the  prism  being  equal  to  E,  we 
have  (VII.,  Proposition  II.) 

s  =  pX  E, 

no  matter  what  the  value  of  n.  If  n  is  indefinitely  increased, 
s  approaches  the  limit  JS  (Proposition  I.),  and  p  X  E,  the 
limit  P  X  E.    Therefore,  by  III.,  Theorem  of  Limits, 

S  =  P  X  E. 


9.  Corollary  I.  The  lateral  area  of  a  cylinder  of  revolution 
is  equal  to  the  product  of  the  circumference  of  its  base  by  its 
altitude. 

This  may  be  formulated, 


U 


if  R  is  the  radius  of  the  base  and  H  the  altitude 

10.  Corollary  II.    The  lateral  ^  ^ 

areas  of  similar  cylinders  of  revo- 
lution are  to  each  other  as  the 
squares  of  their  altitudes,  or  as  the 
squares  of  the  radii  of  their  bases. 

S   _      2t.R.H 

s 


Suggestion. 


R    H 

r  '  h 
by  (5). 


^ 
r* 


jff» 


=  --  =  -—,  smce  -  = 


27:r.h 
R      H 


25* 


h<^ 


294 


ELEMENTS   OP   GEOMETRY. 


PBOPOSITION  III.— THEOREM.  , 

11.  The  volume  of  a  cylinder  is  equal  to  the  product  of  its 
base  by  its  altitude. 

Let  the  volume  of  the  cylinder  be 
denoted  by  F,  its  base  by  B,  and  its 
altitude  by  S.  Let  the  volume  of  an 
inscribed  prism  be  denoted  by  F',  and 
its  base  by  J5';  its  altitude  will  also  be 
-H,  and  we  shall  have  (VII.,  Proposi- 
tion XIL,  Corollary) 

no  matter  what  the  number  of  faces  of  the  prism. 

If  the  number  of  faces  of  the  prism  is  indefinitely  in- 
creased, F'  has  the  limit  F,  and  ^'  X  -S"  the  limit  B  X  S, 
Therefore 

v  =  bxb:. 

12.  Corollary  I.  For  a  cylinder  of  revolution  this  proposition 
may  be  formulated^  V  =  izB^.H.  (Y.,  Proposition  IX.,  Corol- 
lary.) 

13.  Corollary  II.  The  volumes  of  similar  cylinders  of  revo- 
lution are  to  each  other  as  the  cubes  of  their  altitudes^  or  as  the 
cubes  of  their  radii. 


THE  CONE. 

14.  Definition.  The  area  of  the  convex,  or  lateral,  surface 
of  a  cone  is  called  its  lateral  area. 


BOOK   IX. 

15.  Definition.  A  pyramid  is  inscribed 
in  a  cone  when  its  base  is  inscribed  in 
the  base  of  the  cone  and  its  vertex 
coincides  with  the  vertex  of  the  cone. 

It  follows  that  the  lateral  edges  of 
the  pyramid  are  elements  of  the  cone. 

An  inscribed  pyramid  is  wholly  con- 
tained within  the  cone. 

16.  Definition.  A  pyramid  is  circum- 
scribed about  a  cone  when  its  base  is  circumscribed  about  the 
base  of  the  cone  and  its  vertex  co- 
incides with  the  vertex  of  the  cone. 

Any  lateral  face,  as  SAB^  of  the 
pyramid  is  tangent  to  the  cone ;  for, 
since  it  passes  through  a  and  S^  it 
contains  the  element  Sa^  and  it  can- 
not cut  the  convex  surface  again 
without  cutting  the  perimeter  of 
the  base  again  (YIII.,  Proposition 

111.). 

The  cone  is  then  wholly  contained  within  the  pyramid. 

17.  Definition.  A  truncated  cone  is 
the  portion  of  a  cone  included  be- 
tween its  base  and  a  plane  cutting 
its  convex  surface. 

When  the  cutting  plane  is  par- 
allel to  the  base,  the  truncated  cone 
is  called  a  frustum  of  a  cone;  as 
ABCD-abcd.  The  altitude  of  a  frus- 
tum is  the  perpendicular  distance 
Tt  between  its  bases. 

If  a  pyramid  is  inscribed  in  the  cone,  the  cutting  plane 


296 


ELEMENTS   OF   GEOMETRY. 


determines  a  truncated  pyramid  inscribed  in  the  truncated 
cone ;  and  if  a  pyramid  is  circumscribed  about  the  cone,  the 
cutting  plane  determines  a  truncated  pyramid  circumscribed 
about  the  truncated  cone. 

18.  Definition.  In  a  cone  of  revolution 
all  the  elements  are  equal,  and  any  ele- 
ment is  called  the  slant  height  of  the 
cone. 

In  a  cone  of  revolution  the  portion 
of  an  element  included  between  the  par- 
allel bases  of  a  frustum,  as  Aa^  or  Bh^  is 
called  the  slant  height  of  the  frustum. 

19.  Definition.  Similar  cones  of  revolu- 
tion are  those  which  are  generated  by  similar  right  triangles 
revolving  about  homologous  sides. 


PROPOSITION  IV.— THEOREM. 


20.  If  a  'pyramid  he  inscribed  in  or  circumscribed  about  a 
given  cone,  its  volume  will  approach  the  volume  of  the  cone  as  its 
limit,  and  its  lateral  surface  will  approach  the  convex  surface 
of  the  cone  as  a  limit,  as  the  number  of  faces  of  the  pyramid  is 
indefinitely  increased. 

The  demonstration  is  precisely  the  same  as  that  of  Propo- 
sition I.,  substituting  cone  for  cylinder  and  pyramids  for 
prisms. 

21.  Corollary.  A  frustum  of  a  cone  is  the  limit  of  the  in- 
scribed and  circumscribed  frustums  of  pyramids,  the  number  of 
whose  faces  is  indefinitely  increased. 


BOOK   IX. 


297 


PROPOSITION  v.— THEOREM. 

22.  The  lateral  area  of  a  cone  of  revolution  is  equal  to  the 
product  of  the  circumference  of  its  base  by  half  its  slant  height. 

Suggestion.  Circumscribe  a  regular 
pyramid  about  the  cone,  and  then  sup- 
pose the  number  of  its  faces  to  be 
indefinitely  increased,  (y.  YII.,  Prop- 
osition XIY.) 

23.  CoROLLAEY  I.  Thc  proposition 
may  be  formulated,  S  =  tcRI/,  where 
E  is  the  radius  of  the  base  and  L  the 
slant  height. 

24.  Corollary  II.  The  lateral  areas 

of  similar  cones  of  revolution  are  to  each  other  as  the  squares 
of  their  slant  heights,  or  as  the  squares  of  their  altitudes,  or  as 
the  squares  of  the  radii  of  their  bases. 


^^u:;^  PROPOSITION  VI.— THEOREM. 

25.  The  lateral  area  of  a  frustum  of  a  cone  of  revolution  is 
equal  to  the  half  sum  of  the  circumferences  of  its  bases  multi* 
plied  by  its  slant  height. 

Suggestion.  Circumscribe  the  frus- 
tum of  a  regular  pyramid  about  the 
frustum  of  the  cone  (17),  and  sup- 
pose the  number  of  its  faces  indefi- 
nitely increased,  (y.  YII.,  Proposi- 
tion XIY.,  Corollary.) 

26.  Corollary  I.  The  proposition 
may  be  formulated,  S  =  t:{B  -\-  r)L, 
if  R  and  r  are  the  radii  of  the  bases 
and  L  is  the  slant  height. 


298 


ELEMENTS   OF   GEOMETRY. 


27.  Corollary  II.  The  lateral  area  of  a  frustum  of  a  cone 
of  revolution  is  equal  to  the  circumfer- 
ence of  a  section  equidistant  from  its      ] 
bases  multiplied  by  its  slant  height.  jji 

Suggestion,  IKz=  ^(om  +  OM).    (v. 
Exercise  24,  Book  I.) 


PROPOSITION  VII.— THEOREM. 

28.  The  volume  of  any  cone  is  equal  to  one-third  of  the  product 
of  its  base  by  its  altitude. 

Suggestion.  Inscribe  a  pyramid  in 
the  cone,  and  suppose  the  number  of 
its  faces  to  be  indefinitely  increased. 
(v.  VII.,  Proposition  XVIII.) 

29.  Corollary  I.  For  a  cone  of 
revolution,  the  proposition  may  be  for- 
mulated, V  =  InW^.H. 

30.  Corollary  II.  Similar  cones  of 

revolution  are  to  each  other  as  the  cubes  of  their  altitudes,  or  as 
the  cubes  of  the  radii  of  their  bases. 


exercise. 

Theorem. — A  frustum  of  any  cone  is 
equivalent  to  the  sum  of  three  cones  whose 
common  altitude  is  the  altitude  of  the  frus- 
tum, and  whose  bases  are  the  lower  base, 
the  upper  base,  and  a  mean  proportional 
between  the  bases  of  the  frustum,  (v.  VII.,  Proposition  XIX., 
Corollary.) 


BOOK  IX.  299 

THE  SPHERE. 

31.  Definition.  A  spherical  segment  is  a  portion  of  a  sphere 
included  between  two  parallel  planes. 

The  sections  of  the  sphere  made  by  the  parallel  planes  are 
the  bases  of  the  segment ;  the  distance  between  the  planes  is 
the  altitude  of  the  segment. 

Let  the  sphere  be  generated  by  the  revo- 
lution of  the  semicircle  EBF  about  the  axis 
EF;  and  let  Aa  and  Bb  be  two  parallels, 
perpendicular  to  the  axis.  The  solid  gener- 
ated by  the  figure  ABba  is  a  spherical  seg- 
ment ;  the  circles  generated  by  Aa  and  Bb 
are  its  bases ;  and  ab  is  its  altitude. 

If  two  parallels  Aa  and  TE  are  taken,  one 
of  which  is  a  tangent  at  E^  the  solid  generated  by  the  figure 
EAa  is  a  spherical  segment  having  but  one  base,  which  is  the 
section  generated  by  Aa.  The  segment  is  still  included  be- 
tween two  parallel  planes,  one  of  which  is  the  tangent  plane 
at  E,  generated  by  the  line  ET. 

32.  Definition.  A  zone  is  a  portion  of  the  surface  of  a 
sphere  included  between  two  parallel  planes. 

The  circumferences  of  the  sections  of  the  sphere  made  by 
the  parallel  planes  are  the  bases  of  the  zone ;  the  distance 
between  the  planes  is  its  altitude. 

A  zone  is  the  curved  surface  of  a  spherical  segment. 

In  the  revolution  of  the  semicircle  EBF  about  EFy  an  arc 
AB  generates  a  zone ;  the  points  A  and  B  generate  the  bases 
of  the  zone ;  and  the  altitude  of  the  zone  is  ab. 

An  arc,  EA,  one  extremity  of  which  is  in  the  axis,  gener- 
ates a  zone  of  one  base,  which  is  the  circumference  described 
by  the  extremity  A. 


300 


ELEMENTS  OF  GEOMETEY. 


33.  Definition.  When  a  semicircle  revolves  about  its  diam- 
eter, the  solid  generated  by  any  sector  of  the  semicircle  is 
called  a  spherical  sector. 

Thus,  when  the  semicircle  EBF  revolves 
about  EFj  the  circular  sector  COD  generates 
a  spherical  sector. 

The  spherical  sector  is  bounded  by  three 
curved  surfaces;  namely,  the  two  conical 
surfaces  generated  by  the  radii  00  and  OD, 
and  the  zone  generated  by  the  arc  CD.  This 
zone  is  called  the  base  of  the  spherical  sector. 

OD  may,  however,  coincide  with  OF,  in  which  case  the 
spherical  sector  is  bounded  by  a  conical  surface  and  a  zone 
of  one  base. 

-Again,  00  may  be  perpendicular  to  OF,  in  which  case  the 
spherical  sector  is  bounded  by  a  plane,  a  conical  surface,  and 
a  zone. 


V 


PROPOSITION  VIII.— LEMMA. 


34.  The  area  of  the  surface  generated  by  a  straight  line  re- 
volving about  an  axis  in  its  plane,  is  equal  to  the  projection  of 
the  line  on  the  axis  multiplied  by  the  circumference  of  the  circle 
whose  radius  is  the  perpendicular  erected  at  the  middle  of  the 
line  and  terminated  by  the  axis. 

Let  AB  be  the  straight  line  revolving 
about  the  axis  J^Y;  ab  its  projection  on 
the  axis ;  01  the  perpendicular  to  it,  at  its 
middle  point  7,  terminating  in  the  axis; 
then  area  AB  =  ab  X  circ.  01. 

For,  draw  IK  perpendicular  and  AH 
parallel  to  the  axis.    The  area  generated 


BOOK   IX.  301 

by  AB  is  that  of  the  frustum  of  a  cone ;  hence  (Proposition  " 
VI.,  Corollary  II.) 

area  AB  =  AB  X  circ.  IK. 

The  triangles  ABH  and  lOK  are  sinxjlar,  being  mutually 
equiangular,  and  we  have  rCj-^]   / 


but 


an4 


AH^IK       ab_^IK, 
AB        or      AB       OV 

circ^  =  ^(Y.,  Proposition  YIIL), 

ah  circ.  IK 

A  B^  circ.OI' 

ah  X  circ.  01  =  AB  X  circ.  IK. 


au  ;x^  cue.  ui  =  jljd  ;<,  circ.  in..  q 

Therefore  _^--^  ~~^  Jp      V'-'x 

area  AB  =  ah  X  circ.  01  AN     \     ' ' 

If  AB  meets  XY^  the  surface  generated  is  a  conical  sur- 
face ;  but  the  proposition  still  holds,  as  may  be  easily  proved. 
(v.  Proposition  Y.) 

If  AB  is  parallel  to  the  axis,  the  result  is  the  same.  (y. 
Proposition  II.,  Corollary  I.) 


PEOPOSITION  IX.— THEOREM.  ^ 

35.  The  area  of  a  zone  is  equal  to  the  product  of  its  altitude 
by  the  circumference  of  a  great  circle. 

E 

Let  the  sphere  be  generated  by  the  revo-  ^^.^^ 

lution  of  the  semicircle  EBF  about  the  axis  /(i 

EF ;  and  let  the  arc  AD  generate  the  zone  b/^-— X^ 

whose  area  is  required.  I 

Jj^t  the  arc  AD  be  divided  into  any  num-  ^V' 

ber  of  equal  parts.  AB^  BG,  CD,  and  draw  ^^— 
the  chords  AB,  BC,  etc.     These  chords  are 

26 


302 


ELEMENTS   OF   GEOMETRY. 


all  equal,  since  they  subtend  equal  ares ;  and  the  perpendic- 
ulars at  their  middle  points  all  pass  through 
the  centre  0  of  the  semicircle,  and  are  equal 
(II.,  Proposition  YII.). 

Let  abj  he,  etc.,  be  the  projections  of  these 
chords  on  the  axis.  Then,  by  Proposition 
YIII., 

area  AB  =  ab  X  cire.  01, 
area  BC  ==  be  X  cire.  01, 
area  CD  =z  ed  X  cire.  01. 

Hence  the  sum  of  these  areas,  which  is  the  area  generated 
by  the  broken  line  ABCD,  is  equal  to 

(ab  +  6c  +  cd)  X  cire.  01; 

that  is,  to  ad  X  cire.  01. 

Calling  the  area  generated  by  the  broken  inscribed  line,  8y 

we  have 

S  =  ad  X  cire.  01, 

no  matter  what  the  number  of  the  equal  parts  into  which  the 
arc  AD  is  divided.  If,  now,  we  increase  the  number  of  parts 
indefinitely,  01  will  approach  the  radius  of  the  sphere,  and 
cire.  01  the  circumference  of  a  great  circle  as  its  limit,  and 
S  will  approach  the  surface  of  the  zone  as  its  limit.  There* 
fore 

surfaee  of  zone  =  ad  X  cireumferenee  of  great  eirele. 

36.  Corollary.  The  proposition  may  be  formulated, 
S  =  2tzR.H, 


where  R  is  the  radius  of  the  sphere  and  II  Ihe  altitude  of 
the  zone. 


A'-  cJ-r  K'J+ 


t 


BOOK  IX.  303 

PEOPOSITION  X.— THEOREM. 


^       Sfr.  The  area  of  the  surface  of  a  sphere  is  equal  to  the  product 
of  its  diameter  by  the  circumference  of  a  great  circle% 

This  follows  directly  from  Proposition  IX.,  since  the  sur« 
face  of  the  whole  sphere  may  be  regarded  as  a  zone  whose 
altitude  is  the  diameter  of  the  sphere. 

38.  Corollary  I.  This  may  be  formulated^ 

S  =  27: B  X2R  =  4:nR\ 

._       Hence  the  surface  of  a  sphere  is  equivalent  to  four  great  circles.  ' 

39.  Corollary  II.  The  surfaces  of  two  spheres  are  to  each 
other  as  the  squares  of  their  diameters^  or  as  the  squares  of  their 
radii. 

40.  Scholium.  The  area  of  a  spherical  degree  on  a  sphere 

whose  radius  is  R  is  -^  (^IH.,  69),  and,  by  the  aid  of  this 
value,  we  may  readily  reduce  the  area  of  a  spherical  polygon 
to  ordinary  square  measure.  . 


PROPOSITION  XI.— THEOREM. 

41.  The  volume  of  a  sphere  is  equal  to  the  area  of  its  surface 
multiplied  by  one-third  of  its  radius. 

Circumscribe  a  polyedron  about 
the  sphere.  This  may  be  done  by 
taking  at  pleasure  points  on  the  sur- 
face of  the  sphere,  and  drawing 
tangent  planes  at  these  points.  The 
circumscribed  polyedron  wholly  con- 
tains the  sphere,  and  is  greater  than 

the  sphere.  Join  all  the  vertices  of  the  polyedron  with  the 
centre  of  the  sphere,  and  pass  planes  through  the  edges  of 
the  polyedron  and  these  lines,  thus  dividing  the  polyedron 


304  ELEMENTS  OF  GEOMETRY. 

into  pyramids,  each  of  which  has  its  vertex  at  the  centre  of 
the  sphere,  and  has  a  face  of  the  poly- 
edron  as  its  base,  and  has,  therefore, 
the  radius  of  the  sphere  for  its  alti- 
tude (YIII.,  Proposition  YII.).    The 

volume  of  any  one  of  these  pyramids         ^  J^       \pJh 

is  then  one-third  of  the  product  of  a 
face  of  the  polyedron  by  the  radius 
of  the  sphere,  and  the  sum  of  the 

volumes  of  the  pyramids,  or  the  whole  volume  of  the  poly- 
edron, is  one-third  of  the  product  of  the  sum  of  the  faces 
of  the  polyedron  by  the  radius  of  its  sphere ;  that  is,  one-third 
of  the  product  of  the  whole  surface  of  the  polyedron  by  the 
radius  of  the  sphere.  Representing  the  surface  of  the  poly- 
edron by  s,  and  its  volume  by  u,  we  have 

V  =  IBs, 
and  this  equation  holds  no  matter  what  the  number  of  the 
faces  of  the  polyedron. 

If,  now,  we  increase  the  number  of  faces  of  the  polyedron 
by  drawing  additional  tangent  planes  to  the  sphere,  we  de- 
crease the  volume  v,  for  each  new  tangent  plane  cuts  off  a 
corner  of  the  polyedron.  We  may  carry  on  indefinitely  this 
process  of  shaving  down  the  polyedron,  and  may  thus  make 
the  difference  between  its  volume  and  the  volume  of  the 
sphere  as  small  as  we  please ;  but  we  cannot  make  the  two 
volumes  absolutely  coincide.  As  the  two  volumes  approach 
coincidence,  the  two  surfaces  also  approach  coincidence.  If, 
now,  S  is  the  surface  and  V  the  volume  of  the  sphere,  S  is  the 
limit  of  s  and  Y  the  limit  of  v,  as  the  number  of  faces  of  the 
circumscribed  polyedron  is  indefinitely  increased.  Therefore, 
by  III.,  Theorem  of  Limits, 


BOOK  IX.  305 

42.  Corollary  I.  The  result  of  this  proposition  may  be  for- 


mulatedj 


F  =  p' 


43.  Corollary  II.  The  volumes  of  two  spheres  are  to  each 
other  as  the  cubes  of  their  radii,  or  as  the  cubes  of  their  diameters. 


PROPOSITION  XII.— THEOREM. 

44.  The  volume  of  a  spherical  sector  is  equal  to  the  area  of 
the  zone  which  forms  its  base  multiplied  by  one-third  the  radius 
of  the  sphere. 

The  proof  is  analogous  to  the  proof  of  Proposition  XI. 
The  form  of  the  circumscribed  polyedron  is,  however,  some- 
what more  complicated,  as  it  will  be  bounded  by  a  surface 
made  up  of  plane  faces  tangent  to  the  zone  of  the  spherical 
sector,  and  by  two  pyramidal  faces  tangent  to,  or  inscribed  in, 
the  two  conical  surfaces  of  the  spherical  sector. 

45.  Definition.  A  spherical  pyramid  is  a  f. ^^ 

solid  bounded  by  a  spherical  polygon  and         y^  \        /\ 
the  planes  of  the  sides  of  the  polygon ;  as      ^  \       /\,''' 
0-ABCD.    The  centre  of  the  sphere  is  the  \  //  \    / 

vertex  of  the  pyramid  j  the  spherical  poly-        •      '^~ Y 

gon  is  its  base. 


EXERCISE. 

Theorem. —  The  volume  of  a  spherical  pyramid  is  equal  to  the 
area  of  its  base  multiplied  by  one-third  of  the  radius  of  the 
sphere. 

u  26* 


306 


ELEMENTS   OF   GEOMETRY. 


y 


PROPOSITION  XIII.— PROBLEM. 

46.  To  find  the  volume  of  a  spherical  segment. 

Any  spherical  segment  may  be  obtained  from  a  spherical 
sector  by  adding  to  it,  or  subtracting  from  it,  cones  having  as 
bases  the  bases  of  the  segment. 

For  example,  let  us  consider  a  segment  of 
two  bases  which  does  not  contain  the  centre 
of  the  sphere.  The  segment  generated  by 
the  revolution  of  ABGD  about  OC  may  be  ' 
obtained  by  taking  the  cone  generated  by 
OAD  from  the  sum  of  the  cone  generated  by 
OBG  and  the  spherical  sector  generated  by  OAB. 

Call  OC  /,  ODp,  DC  h,  AD  r,  BC  r',  and  OA  E,  and  the 
volume  of  the  segment  F.    Then  we  have  the  simple  relations 


h 

r'  +  p' 


P  —Pi 
K*,  r'^  +  p"" 


B\ 


The  area  of  the  zone  of  the  segment  is  27ri2.^  (Proposition 
IX.,  Corollary).     Hence      ,  ^  ^^/{ 

V  =  ^T:hB^  +  i7r/r"-—  Upr"  (Proposition  XII.,  and  Proposi- 

^ -  .T^~--^:;r2r:^^^  tion  YII.,  Corollayg  I.), 

y  =  !;:(/  -"^  +  %^^^i§^'^i  -  i^p(^  -%^ 

V  =  (/  -  p)7:R'  -  UG"  -^  p%  [1] 

a  convenient  formula  when  the  distances  of  the  bases  of  the 
segment  from  the  centre  of  the  sphere  are  given. 

Another  convenient  formula  can  be  obtained  by  introducing 
in  [1]  A,  r,  and  /  in  place  of  j?  and  p'.    We  have 


F=(/-;>)|[3iJ' 


(y  +  /j'  +  i'0]- 


BOOK   IX. 


Now 
Hence 

and 


2/i>  +  ;>'- 


,;^it+Jt^\ 


=  f  (iP  - /' +  iJ' -  r")  - 1  =  3JP  -  Kr"  + /^  - 1', 


and  we  have 


[2] 


This  formula  is  convenient  when  the  areas  of  the  bases  of 
the  segment  are  given,  and  it  may  be  put  into  words  as 
follows : 

The  volume  of  a  spherical  segment  is  equal  to  the  half  sum 
of  its  bases  multiplied  by  its  altitude  plus  the  volume  of  a  sphere 
of  which  that  altitude  is  the  diameter,  . 


V^^ 


\  -.^^; 


EXERCISES  ON  BOOK  IX. 


X 


THEOREMS. 


1.  Give  a  strict  proof  of  Proposition  I.  and  Proposition  IV.  for 
the  volumes  of  cylinder  and  cone,  by  showing  that  the  difference 
between  the  volumes  of  the  inscribed  and  circumscribed  figures 
can  be  decreased  at  pleasure. 

2.  Assuming  that  if  a  solid  has  a  plane  face  the  area  of  that  face 
is  less  than  the  rest  of  the  surface  of  the  solid,  prove,  first,  that 
if  two  convex  solids  have  a  plane  face  in  common,  and  one  solid 
is  wholly  included  by  the  other,  its  surface  is  less  than  that  of  the 
other  {v.  V.,  13),  and  then  give  a  strict  proof  of  Proposition  I.  and 
Proposition  IV.  for  the  surfaces  of  cylinder  and  cone. 

8.  The  volumes  of  a  cone  of  revolution,  a  sphere,  and  a  cylinder 
of  revolution  are  proportional  to  the  numbers  1,  2,  3  if  the  bases 
of  the  cone  and  cylinder  are  each  equal  to  a  great  circle  of  the 
sphere,  and  their  altitudes  are  each  equal  to  a  diameter  of  the 
sphere. 

4.  An  equilateral  cylinder  (of  revolution)  is  one  a  section  of 
which  through  the  axis  is  a  square.  An  equilateral  cone  (of 
revolution)  is  one  a  section  of  which  through  the  axis  is  an  equi- 
lateral triangle.  These  definitions  premised,  prove  the  following 
theorems : 

I.  The  total  area  of  the  equilateral  cylinder  inscribed  in  a 
sphere  is  a  mean  proportional  between  the  area  of  the  sphere  and 
the  total  area  of  the  inscribed  equilateral  cone.  The  same  is  true 
of  the  volumes  of  these  three  bodies. 

II.  The  total  area  of  the  equilateral  cylinder  circumscribed 
about  a  sphere  is  a  mean  proportional  between  the  area  of  the 
sphere  and  the  total  area  of  the  circumscribed  equilateral  cone. 
The  same  is  true  of  the  volumes  of  these  three  bodies. 

5.  If  h  is  the  altitude  of  a  segment  of  one  base  in  a  sphere  whose 
radius  is  r,  the  volume  of  the  segment  is  equal  to  Trh'^H  —  ^h), 

6.  The  volumes  of  polyedrons  circumscribed  about  the  same 
sphere  are  proportional  to  their  surfaces. 

308  . 

7^ 


MISCELLANEOUS  EXERCISES 

ON   THE 

GEOMETKY  OF  SPACE. 


1.  A  PERPENDICULAR  let  fall  from  the  middle  point  of  a  line 
upon  any  plane  not  cutting  the  line  is  equal  to  one-half  the  sum 
of  the  perpendiculars  let  fall  from  the  ends  of  the  line  upon  the 
same  plane. 

2.  The  perpendicular  let  fall  from  the  point  of  intersection  of 
the  medial  lines  of  a  given  triangle  upon  any  plane  not  cutting 
the  triangle  is  equal  to  one-third  the  sum  of  the  perpendiculars 
from  the  vertices  of  the  triangle  upon  the  same  plane. 

3.  The  perpendicular  from  the  centre  of  gravity  of  a  tetraedron 
upon  any  plane  not  cutting  the  tetraedron  is  equal  to  one-fourth 
the  sum  of  the  perpendiculars  from  the  vertices  of  the  tetraedron 
upon  the  same  plane. 

4.  The  volume  of  a  truncated  triangular  prism  is  equal  to  the 
product  of  the  area  of  its  lower  base  by  the  perpendicular  upon 
the  lower  base  let  fall  from  the  intersection  of  the  medial  lines  of 
the  upper  base. 

6.  The  volume  of  a  truncated  parallelopiped  is  equal  to  the 
product  of  the  area  of  its  lower  base  by  the  perpendicular  from 
the  centre  of  the  upper  base  upon  the  lower  base. 

6.  If  ABCD  is  any  tetraedron,  and  O  any  point  within  it,  and 
if  the  straight  lines  AO^  BO^  CO^  DO,  are  produced  to  meet  the 
faces  in  the  points  a,  6,  c,  d,  respectively,  then 


Aa'^  Bb~^  Cc  "^  Dd 


809 


310  -  ELEMENTS   OF   GEOMETRY. 

/7.  If  the  three  face  angles  of  the  vertical  triedral  angle  of  a 
tetraedron  are  right  angles,  and  the  lengths  of  the  lateral  edges 
are  represented  by  a,  6,  and  c,  and  of  the  altitude  by  p,  then 

8.  If  the  three  face  angles  of  the  vertical  triedral  angle  of  a 
tetraedron  are  right  angles,  the  square  of  the  area  of  the  base  is 
equal  to  the  sum  of  the  squares  of  the  areas  of  the  lateral  faces. 

9.  The  perpendicular  from  the  middle  point  of  the  diagonal  of 
a  rectangular  parallelepiped  upon  a  lateral  edge  bisects  the  edge, 
and  is  equal  to  one-half  of  the  projection  of  the  diagonal  upon 
the  base. 

10.  A  straight  line  of  a  given  length  moves  so  that  its  extremi- 
ties are  constantly  upon  two  given  perpendicular  but  non-inter- 
secting straight  lines :  what  is  the  locus  of  the  middle  point  of  the 
moving  line  ? 


PROBLEMS. 


11.  To  cut  a  given  polyedral  angle  of  four  fac^s  by  a  plane  so 
that  the  section  shall  be  a  parallelogram. 

^\    12.  To  cut  a  cube  by  a  plane  so  that  the  section  shall  be  a  regular 
hexagon. 

13.  To  find  the  ratio  of  the  volumes  generated  by  a  rectangle 
revolving  successively  about  its  two  adjacent  sides. 


SYLLABUS  OP  PROPOSITIONS 


IN 


SOLID     GEOMETRY. 


BOOK  VI. 

THEOREMS. 

Proposition  I. 

Through  any  given  straight  line  a  plane  may  be  passed,  but 
the  line  will  not  determine  the  plane. 

Proposition  II. 

A  plane  is  determined,  1st,  by  a  straight  line  and  a  point  with- 
out that  line  ;  2d,  by  two  intersecting  straight  lines  ;  3d,  by  three 
points  not  in  the  same  straight  line  ;  4th,  by  two  parallel  straight 
lines. 

Corollary,  The  intersection  of  two  planes  is  a  straight  line. 

Proposition  III. 

From  a  given  point  without  a  plane  one  perpendicular  to  the 
plane  can  be  drawn,  and  but  one ;  and  the  perpendicular  is  the 
shortest  line  that  can  be  drawn  from  the  point  to  the  plane. 

Corollary.  At  a  given  point  in  a  plane  one  perpendicular  can  be 
erected  to  the  plane,  and  but  one. 

Proposition  IV. 

If  a  straight  line  is  perpendicular  to  each  of  two  straight  lines 
at  their  point  of  intersection,  it  is  perpendicular  to  the  plane  of 
those  lines. 

Corollary  I.  At  a  given  point  of  a  straight  line,  one  plane  can 
be  drawn  perpendicular  to  the  line,  and  but  one. 

Corollary  II.  Through  a  given  point  without  a  straight  line, 
one  plane  can  be  drawn  perpendicular  to  the  line,  and  but  one. 

311 


312  ELEMENTS   OF   GEOMETRY. 


Proposition  V. 

Two  lines  in  space  having  the  same  direction  are  parallel. 
Corollary,  Two  lines  parallel  to  the  same  line  are  parallel  to 
each  other. 

Proposition  VI. 

If  two  straight  lines  are  parallel,  every  plane  passed  through 
one  of  them  and  not  coincident  with  the  plane  of  the  parallels  is 
parallel  to  the  other. 

Corollary  I.  Through  any  given  straight  line  a  plane  can  be 
passed  parallel  to  any  other  given  straight  line. 

Corollary  II.  Through  any  given  point  a  plane  can  be  passed 
parallel  to  any  two  given  straight  lines  in  space. 

Proposition  VII. 

Planes  perpendicular  to  the  same  straight  line  are  parallel  to 
each  other. 

Proposition  VIII. 

The  intersections  of  two  parallel  planes  with  any  third  plane 
are  parallel. 

Proposition  IX. 

A  straight  line  perpendicular  to  one  of  two  parallel  planes  is 
perpendicular  to  the  other. 

Corollary.  Through  any  given  point  one  plane  can  be  passed 
parallel  to  a  given  plane,  and  but  one. 

Proposition  X. 

If  two  angles,  not  in  the  same  plane,  have  their  sides  respec- 
tively parallel  and  lying  in  the  same  direction,  they  are  equal 
and  their  planes  are  parallel. 

Proposition  XI. 

If  one  of  two  parallel  lines  is  perpendicular  to  a  plane,  the 
other  is  also  perpendicular  to  that  plane. 

Corollary.  Two  straight  lines  perpendicular  to  the  same  plane 
are  parallel  to  each  other. 

Proposition  XII. 
Two  diedral  angles  are  equal  if  their  plane  angles  are  equal. 

Proposition  XIII. 
Two  diedral  angles  are  in  the  same  ratio  as  their  plane  angles. 


SYLLABUS   OF   PROPOSITIONS   IN   SOLID   GEOMETRY.       313 

Proposition  XIV. 

If  a  straight  line  is  perpendicular  to  a  plane,  every  plane  passed 
through  the  line  is  perpendicular  to  the  plane. 

Proposition  XV. 

If  two  planes  are  perpendicular  to  each  other,  a  straight  line 
drawn  in  one  of  them,  perpendicular  to  their  intersection,  is 
perpendicular  to  the  other. 

Corollary  I.  If  two  planes  are  perpendicular  to  each  other,  a 
straight  line  drawn  through  any  point  of  their  intersection  per- 
pendicular to  one  of  the  planes  will  lie  in  the  other. 

Corollary  II.  If  two  planes  are  perpendicular,  a  straight  line 
let  fall  from  any  point  of  one  plane  perpendicular  to  the  other 
will  lie  in  the  first  plane. 

Proposition  XVI. 

If  two  intersecting  planes  are  each  perpendicular  to  a  third 
plane,  their  intersection  is  also  perpendicular  to  that  plane. 

Proposition  XVII. 

Through  any  given  straight  line  a  plane  can  be  passed  perpen- 
dicular to  any  given  plane. 

Proposition  XVIII. 
The  projection  of  a  straight  line  upon  a  plane  is  a  straight  line. 

Proposition  XIX. 

The  acute  angle  which  a  straight  line  makes  with  its  own  pro- 
jection upon  a  plane  is  the  least  angle  it  makes  with  any  line  of 
that  plane. 

Proposition  XX. 

The  sum  of  any  two  face  angles  of  a  triedral  angle  is  greater 
than  the  third. 

Proposition  XXI. 

The  sum  of  the  face  angles  of  any  convex  polyedral  angle  is 
less  than  four  right  angles. 

Proposition  XXII. 

If  two  triedral  angles  have  the  three  face  angles  of  the  one  re- 
spectively equal  to  the  three  face  angles  of  the  other,  the  corre- 
sponding diedral  angles  are  equal. 
0  27 


314  ELEMENTS   OF   GEOMETRY. 

BOOK  VII. 

THEOREMS. 
Proposition  I. 

The  sections  of  a  prism  made  by  parallel  planes  are  equal  poly- 
gons. 

Corollary.  Any  section  of  a  prism  made  by  a  plane  parallel  to 
the  base  is  equal  to  the  base. 

Proposition  II. 

The  lateral  area  of  a  prism  is  equal  to  the  product  of  the  perim- 
eter of  a  right  section  of  the  prism  by  a  lateral  edge. 

Corollary.  The  lateral  area  of  a  right  prism  is  equal  to  the 
product  of  the  perimeter  of  its  base  by  its  altitude. 

Proposition  III. 

Two  prisms  are  equal,  if  three  faces  including  a  triedral  angle 
of  the  one  are  respectively  equal  to  three  faces  similarly  placed 
including  a  triedral  angle  of  the  other. 

Corollary  I.  Two  truncated  prisms  are  equal,  if  three  faces  in- 
cluding a  triedral  angle  of  the  one  are  respectively  equal  to  three 
faces  similarly  placed  including  a  triedral  angle  of  the  other. 

Corollary  II.  Two  ri^ht  prisms  are  equal  if  they  have  equal 
bases  and  equal  altitudes. 

Proposition  IV. 

Any  oblique  prism  is  equivalent  to  a  right  prism  whose  base  is 
a  right  section  of  the  oblique  prism,  and  whose  altitude  is  equal 
to  a  lateral  edge  of  the  oblique  prism. 

Proposition  V. 

Any  parallelopiped  is  equivalent  to  a  rectangular  parallelopiped 
of  the  same  altitude  and  an  equivalent  base. 

Proposition  VI. 

The  plane  passed  through  two  diagonally  opposite  edges  of  a 
parallelopii^ed  divides  it  into  two  equivalent  triangular  prisms. 

Proposition  VII. 

Two  rectangular  parallelepipeds  having  equal  bases  are  to  each 
other  as  their  altitudes. 


SYLLABUS   OF   PROPOSITIONS   IN   SOLID   GEOMETRY.      315 

Proposition  VIII. 

Two  rectangular  parallelepipeds  having  equal  altitudes  are  to 
each  other  as  their  bases. 

Proposition  IX. 

Any  two  rectangular  parallelopipeds  are  to  each  other  as  the 
products  of  their  three  dimensions. 

Proposition  X. 

The  volume  of  a  rectangular  parallelopiped  is  equal  to  tne 
product  of  its  three  dimensions,  the  unit  of  volume  being  the 
cube  whose  edge  is  the  linear  unit. 

Proposition  XI. 

The  volume  of  any  parallelopiped  is  equal  to  the  product  of  the 
area  of  its  base  by  its  altitude. 

Proposition  XII. 

The  volume  of  a  triangular  prism  is  equal  to  the  product  of  its 
base  by  its  altitude. 

Corollary.  The  volume  of  any  prism  is  equal  to  the  product  of 
its  base  by  its  altitude. 

Proposition  XIII. 

If  a  pyramid  is  cut  by  a  plane  parallel  to  its  base,  1st,  the  edges 
and  the  altitude  are  divided  proportionally ;  2d,  the  section  is  a 
polygon  similar  to  the  base. 

Corollary  I.  If  a  pyramid  is  cut  by  a  plane  parallel  to  its  base, 
the  area  of  the  section  is  to  the  area  of  the  base  as  the  square  of 
its  distance  from  the  vertex  is  to  the  square  of  the  altitude  of  the 
pyramid. 

Corollary  II.  If  two  pyramids  have  equal  altitudes  and  equiva- 
lent bases,  sections  made  by  planes  parallel  to  their  bases  and  at 
equal  distances  from  their  vertices  are  equivalent. 

Proposition  XIV. 

The  lateral  area  of  a  regular  pyramid  is  equal  to  the  product  of 
the  perimeter  of  its  base  by  half  its  slant  height. 

Corollary.  The  lateral  area  of  the  frustum  of  a  regular  pyra- 
mid is  equal  to  the  half  sum  of  the  perimeters  of  its  bases  multi- 
plied by  the  slant  height  of  the  frustum. 


316  ELEMENTS   OF   GEOMETRY. 

Proposition  XV. 

If  the  altitude  of  any  given  triangular  pyramid  is  divided  into 
equal  parts,  and  through  the  points  of  division  planes  are  passed 
parallel  to  the  base  of  the  pyramid,  and  on  the  sections  made  by 
these  planes  as  upper  bases  prisms  are  described  having  their 
edges  parallel  to  an  edge  of  the  pyramid  and  their  altitudes  equal 
to  one  of  the  equal  parts  into  which  the  altitude  of  the  pyramid 
is  divided,  the  total  volume  of  these  prisms  will  approach  the 
volume  of  the  pyramid  as  its  limit  as  the  number  of  parts  into 
which  the  altitude  of  the  pyramid  is  divided  is  indefinitely 
increased. 

Proposition  XVI. 

Two  triangular  pyramids  having  equivalent  bases  and  equal 
altitudes  are  equivalent. 

Proposition  XVII. 

A  triangular  pyramid  is  one-third  of  a  triangular  prism  of  the 
same  base  and  altitude. 

Corollary.  The  volume  of  a  triangular  pyramid  is  equal  to  one- 
third  of  the  product  of  its  base  by  its  altitude. 

Proposition  XVIII. 

The  volume  of  any  pyramid  is  equal  to  one-third  of  the  product 
of  its  base  by  its  altitude. 

Proposition  XIX. 

A  frustum  of  a  triangular  pyramid  is  equivalent  to  the  sum  of 
three  pyramids  whose  common  altitude  is  the  altitude  of  the  frus- 
tum, and  whose  bases  are  the  lower  base,  the  upper  base,  and  a 
mean  proportional  between  the  bases  of  the  frustum. 

Corollary.  A  frustum  of  any  pyramid  is  equivalent  to  the  sum 
of  three  pyramids  whose  common  altitude  is  the  altitude  of  the 
frustum,  and  whose  bases  are  the  lower  base,  the  upper  base,  and 
a  mean  proportional  between  the  bases  of  the  frustum. 

Proposition  XX. 

A  truncated  triangular  prism  is  equivalent  to  the  sum  of  three 
pyramids  whose  common  base  is  the  base  of  the  prism  and  whose 
vertices  are  the  three  vertices  of  the  inclined  section. 

Proposition  XXI. 
Only  five  regular  (convex)  polyedrons  are  possible. 


SYLLABUS  OF   PROPOSITIONS   IN  SOLID   GEOMETRY.       317 

BOOK  Till. 
THEOREMS. 
Proposition  I. 

Every  section  of  a  cylinder  made  by  a  plane  passing  through 
an  element  is  a  parallelogram. 

Corollary.  Every  section  of  a  right  cylinder  made  by  a  plane 
perpendicular  to  its  base  is  a  rectangle. 

Proposition  II. 

The  bases  of  a  cylinder  are  equal. 

Corollary  I.  Any  two  parallel  sections  of  a  cylindrical  surface 
are  equal. 

Corollary  II.  All  the  sections  of  a  circular  cylinder  parallel  to 
its  bases  are  equal  circles,  and  the  straight  line  joining  the  centres 
of  the  bases  passes  through  the  centres  of  all  the  parallel  sections. 

Proposition  III. 

Every  section  of  a  cone  made  by  a  plane  passing  through  its 
vertex  is  a  triangle. 

Proposition  IV. 

If  the  base  of  a  cone  is  a  circle,  every  section  made  by  a  plane 
parallel  to  the  base  is  a  circle. 

Corollary.  The  axis  of  a  circular  cone  passes  through  the  centres 
of  all  the  sections  parallel  to  the  base. 

Proposition  V. 

Every  section  of  a  sphere  made  by  a  plane  is  a  circle. 

Corollary  I.  The  axis  of  a  circle  on  a  sphere  passes  through 
the  centre  of  the  circle. 

Corollary  II.  All  great  circles  of  the  same  sphere  are  equal. 

Corollary  III.  Every  great  circle  divides  the  sphere  into  two 
equal  parts. 

Corollary  IV.  Any  two  great  circles  on  the  same  sphere  bisect 
each  other. 

Corollary  V.  An  arc  of  a  great  circle  may  be  drawn  through  any 
two  given  points  on  the  surface  of  a  sphere,  and,  unless  the  points 
are  the  opposite  extremities  of  a  diameter,  only  one  such  arc  can 
be  drawn. 

Corollary  VI.  An  arc  of  a  circle  may  be  drawn  through  any 
three  given  points  on  the  surface  of  a  sphere. 

27* 


318  ELEMENTS   OF   GEOMETRY. 

Proposition  VI. 

All  the  points  in  the  circumference  of  a  circle  on  a  sphere  are 
equally  distant  from  either  of  its  poles. 

Corollary  I.  All  the  arcs  of  great  circles  drawn  from  a  pole  of  a 
circle  to  points  in  its  circumference  are  equal. 

Corollary  II.  The  polar  distance  of  a  great  circle  is  a  quadrant. 

Corollary  III.  If  a  point  on  the  surface  of  a  sphere  is  at  a  quad- 
rant's distance  from  each  of  two  given  points  of  the  surface, 
which  are  not  opposite  extremities  of  a  diameter,  it  is  the  pole 
of  the  great  circle  passing  through  them. 

Proposition  VII. 

A  plane  tangent  to  a  sphere  is  perpendicular  to  the  radius 
drawn  to  the  point  of  contact. 

Corollary.  A  plane  perpendicular  to  a  radius  of  a  sphere  at  its 
extremity  is  tangent  to  the  sphere. 

Proposition  VIII. 

The  intersection  of  two  spheres  is  a  circle  whose  plane  is  per-^ 
pendicular  to  the  straight  line  joining  their  centres,  and  whose 
centre  is  in  that  line. 

Proposition  IX. 

The  angle  of  two  arcs  of  great  circles  is  equal  to  the  angle  of 
their  planes,  and  is  measured  by  the  arc  of  a  great  circle  described 
from  its  vertex  as  a  pole  and  included  between  its  sides  (produced 
if  necessary). 

Corollary.  All  arcs  of  great  circles  drawn  through  the  pole  of  a 
given  great  circle  are  perpendicular  to  its  circumference. 

Proposition  X. 

If  the  first  of  two  spherical  triangles  is  the  polar  triangle  of  the 
second,  then,  reciprocally,  the  second  is  the  polar  triangle  of  the 
first. 

Proposition  XI. 

In  two  polar  triangles,  each  angle  of  one  is  measured  by  the 
supplement  of  the  side  lying  opposite  to  it  in  the  other. 

Proposition  XII. 

Two  triangles  on  the  same  sphere  are  either  equal  or  symmet- 
rical when  two  sides  and  the  included  angle  of  one  are  respectively 
equal  to  two  sides  and  the  included  angle  of  the  other. 


SYLLABUS   OF   PROPOSITIONS   IN   SOLID   GEOMETRY.       319 

Proposition  XIII. 

Two  triangles  on  the  same  sphere  are  either  equal  or  symmet- 
rical when  a  side  and  the  two  adjacent  angles  of  one  are  respec- 
tively equal  to  a  side  and  the  two  adjacent  angles  of  the  other. 

Proposition  XIV. 

Two  triangles  on  the  same  sphere  are  either  equal  or  symmet- 
rical when  the  three  sides  of  one  are  respectively  equal  to  the 
three  sides  of  the  other. 

Proposition  XV. 

If  two  triangles  on  the  same  sphere  are  mutually  equiangular, 
they  are  mutually  equilateral,  and  are  either  equal  or  symmet- 
rical. 

Proposition  XVI. 

Any  side  of  a  spherical  triangle  is  less  than  the  sum  of  the 
other  two. 

Proposition  XVII. 

The  sum  of  the  sides  of  a  convex  spherical  polygon  is  less  than 
the  circumference  of  a  great  circle. 

Proposition  XVIII. 

The  sum  of  the  angles  of  a  spherical  triangle  is  greater  than 
two,  and  less  than  six,  right  angles. 

Proposition  XIX. 
Two  symmetrical  spherical  triangles  are  equivalent. 

Proposition  XX. 

If  two  arcs  of  great  circles  intersect  on  the  surface  of  a  hemi- 
sphere, the  sum  of  the  opposite  spherical  triangles  which  they 
form  is  equivalent  to  a  lune  whose  angle  is  the  angle  between  the 
arcs  in  question. 

Proposition  XXI. 

A  lune  is  to  the  surface  of  the  sphere  as  the  angle  of  the  lune  is 
to  four  right  angles. 

Corollary.  The  area  of  a  lune  is  expressed  by  twice  its  angle, 
the  angular  unit  being  the  degree,  and  the  unit  of  surface  the 
spherical  degree. 


320  ELEMENTS   OF   GEOMETRY. 

Proposition  XXII. 
The  area  of  a  spherical  triangle  is  equal  to  the  excess  of  the 
sum  of  its  angles  over  two  right  angles. 

Proposition  XXIII. 
The  shortest  line  that  can  be  drawn  on  the  surface  of  a  sphere 
between  two  points  is  the  arc  of  a  great  circle,  not  greater  than  a 
semi-circumference,  joining  the  two  points. 


BOOK  IX. 

THEOREMS. 
Proposition  I. 
If  a  prism  whose  base  is  a  regular  polygon  be  inscribed  in  or 
circumscribed  about  a  given  cylinder,  its  volume  will  approach 
the  volume  of  the  cylinder  as  its  limit,  and  its  lateral  surface  will 
approach  the  lateral  surface  of  the  cylinder  as  its  limit  as  the 
number  of  sides  of  its  base  is  indefinitely  increased. 

Proposition  II. 

The  lateral  area  of  a  cylinder  is  equal  to  the  product  of  the 
perimeter  of  a  right  section  of  the  cylinder  by  an  element  of  the 
surface. 

Corollary  I.  The  lateral  area  of  a  cylinder  of  revolution  is  equal 
to  the  product  of  the  circumference  of  its  base  by  its  altitude. 
This  may  be  formulated, 

Corollary  II.  The  lateral  areas  of  similar  cylinders  of  revolu- 
tion are  to  each  other  as  the  squares  of  their  altitudes,  or  as  the 
squares  of  the  radii  of  their  bases. 

Proposition  III. 

The  volume  of  a  cylinder  is  equal  to  the  product  of  its  base  by 
its  altitude. 

Corollary  I.  For  a  cylinder  of  revolution  this  may  be  formu- 
lated, 

Corollary  II.  The  volumes  of  similar  cylinders  of  revolution 
are  to  each  other  as  the  cubes  of  their  altitudes,  or  as  the  cubes 
of  their  radii. 


SYLLABUS  OF   PROPOSITIONS   IN   SOLID  GEOMETRY.       321 

Proposition  IV. 

If  a  pyramid  be  inscribed  in  or  circumscribed  about  a  given 
cone,  its  volume  will  approach  the  volume  of  the  cone  as  its  limit, 
and  its  lateral  surface  will  approach  the  convex  surface  of  the 
cone  as  its  limit  as  the  number  of  faces  of  the  pyramid  is  indefi- 
nitely increased. 

Corollary,  A  frustum  of  a  cone  is  the  limit  of  the  inscribed  and 
circumscribed  frustums  of  pyramids,  the  number  of  whose  faces 
is  indefinitely  increased. 

Proposition  V. 

The  lateral  area  of  a  cone  of  revolution  is  equal  to  the  product 
of  the  circumference  of  its  base  by  half  its  slant  height. 
Corollary  I.  This  proposition  may  be  formulated, 

S  =  ^RL. 

Corollary  II.  The  lateral  areas  of  similar  cones  of  revolution 
are  to  each  other  as  the  squares  of  their  slant  heights,  or  as  the 
squares  of  their  altitudes,  or  as  the  squares  of  the  radii  of  their 
bases. 

Proposition  VI. 

The  lateral  area  of  a  frustum  of  a  cone  of  revolution  is  equal  to 
the  half  sum  of  the  circumferences  of  its  bases  multiplied  by  its 
slant  height. 

Corollary  I.  This  proposition  may  be  formulated, 

S=Tr{R-{.r)L. 

Corollary  II.  The  lateral  area  of  a  frustum  of  a  cone  of  revolu- 
tion is  equal  to  the  circumference  of  a  section  equidistant  from  its 
bases  multiplied  by  its  slant  height. 

Proposition  VII. 

The  volume  of  any  cone  is  equal  to  one-third  the  product  of  its 
base  by  its  altitude. 

Corollary  I.  For  a  cone  of  revolution  this  proposition  may  be 
formulated, 

V=  ITER'S. 

Corollary  II.  Similar  cones  of  revolution  are  to  each  other  aa 
the  cubes  of  their  altitudes,  or  as  the  cubes  of  the  radii  of  their 


322  ELEMENTS  OF  GEOMETRY. 

Proposition  VIII. 

The  area  of  the  surface  generated  by  a  straight  line  revolving 
about  an  axis  in  its  plane  is  equal  to  the  projection  of  the  line 
on  the  axis  multiplied  by  the  circumference  of  the  circle  whose 
radius  is  the  perpendicular  erected  at  the  middle  of  the  line  and 
terminated  by  the  axis. 

Proposition  IX. 

The  area  of  a  zone  is  equal  to  the  product  of  its  altitude  by  the 
circumference  of  a  great  circle. 

Corollary.  This  proposition  may  be  formulated, 

Proposition  X. 

The  area  of  the  surface  of  a  sphere  is  equal  to  the  product  of^ 
its  diameter  by  the  circumference  of  a  great  circle. 
Corollary  I.  This  may  be  formulated, 

S  =  27ri2  X2B  =  ^TzB^,- 

Hence  the  surface  of  a  sphere  is  equivalent  to  four  great  circles. 

Corollary  II.  The  surfaces  of  two  spheres  are  to  each  other  aa 
the  squares  of  their  diameters,  or  as  the  squares  of  their  radii. 

Proposition  XI. 

The  volume  of  a  sphere  is  equal  to  the  area  of  its  surface  multi- 
plied by  one-third  of  its  radius. 
Corollary  I.  This  proposition  may  be  formulated, 

Corollary  II.  The  volumes  of  two  spheres  are  to  each  other  as 
the  cubes  of  their  radii,  or  as  the  cubes  of  their  diameters. 

Proposition  XII. 

The  volume  of  a  spherical  sector  is  equal  to  the  area  of  the  zone 
which  forms  its  base  multiplied  by  one-third  the  radius  of  the 
sphere.  ^__ 

"  Of  THB 


THE  END.  T  ^  I T^  F 


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